Welcome, future physicists! Today, we're going to embark on a fascinating journey into the cosmos and understand how those incredible satellites, both natural (like our Moon) and artificial (like the ones providing your internet and GPS), manage to stay up there, seemingly defying gravity. This topic is fundamental not just for your exams but for understanding the universe around us!

Let's start from scratch and build our intuition step-by-step.

### 1. What Exactly is a Satellite and an Orbit?

Imagine throwing a stone horizontally. It travels some distance and then falls to the ground. If you throw it harder, it goes further before falling. What if you could throw it *really* hard?

This brings us to a brilliant thought experiment by none other than Sir Isaac Newton. Imagine a giant cannon on top of a very high mountain, above the Earth's atmosphere.

* If you fire the cannonball with low speed, it falls to Earth (path A).
* If you fire it with a higher speed, it falls further away (path B).
* If you fire it with an even higher speed, it might even go around the Earth once before falling (path C).
* Now, what if you fire it with *just the right speed*? The cannonball would continuously fall towards the Earth, but because the Earth's surface is curved, it would keep missing the ground! It would literally fall *around* the Earth, never hitting it. This continuous "falling around" is what we call an orbit.

A satellite is essentially any object that orbits another larger object (the central body) due to gravitational attraction. Our Moon is a natural satellite of Earth, and the International Space Station (ISS) is an artificial satellite.

So, the key idea here is: A satellite is constantly falling towards the central body, but its tangential velocity is so high that it never hits the surface. It's a delicate balance between the pull of gravity and the satellite's forward motion.

### 2. The Dance of Forces: Why Orbits are Stable

For a satellite to maintain a stable circular orbit around a central body (like the Earth), two forces must be perfectly balanced:

1. Gravitational Force (F_g): This is the attractive force exerted by the central body (Earth, mass M) on the satellite (mass m). It always pulls the satellite *inwards*, towards the center of the Earth.

We know this force from Newton's Law of Universal Gravitation:

F_g = GMm / r²

where G is the universal gravitational constant, M is the mass of the central body, m is the mass of the satellite, and r is the radius of the orbit (distance from the center of the Earth to the satellite).

2. Centripetal Force (F_c): This is the "apparent" force required to keep an object moving in a circular path. It also acts *inwards*, towards the center of the circle. This force isn't a new kind of force; it's just the net force that causes circular motion. In the case of a satellite, the gravitational force *provides* the necessary centripetal force.

The formula for centripetal force is:

F_c = mv² / r

where m is the mass of the satellite, v is its orbital speed, and r is the radius of the orbit.

For a stable orbit, these two forces must be equal:

F_g = F_c

GMm / r² = mv² / r

This fundamental balance is what allows us to derive all the properties of satellite motion!

### 3. Orbital Speed (v_o): How Fast Does a Satellite Need to Go?

The orbital speed is the tangential speed a satellite must have to maintain a stable orbit at a given radius. Let's derive it from our force balance equation:

GMm / r² = mv² / r

Notice that the mass of the satellite, 'm', cancels out on both sides! This is a super important point:

The orbital speed of a satellite does NOT depend on its own mass!

A tiny screw and the massive International Space Station will have the same orbital speed if they are orbiting at the same radius around the same central body.

Let's simplify the equation:

GM / r = v²

Taking the square root:

v_o = √(GM / r)

Here:
* G is the Universal Gravitational Constant (approx. 6.674 × 10⁻¹¹ N m²/kg²).
* M is the mass of the central body (e.g., Earth's mass = 5.972 × 10²⁴ kg).
* r is the orbital radius (distance from the *center* of the central body to the satellite). Remember, if the altitude 'h' is given, then r = R + h, where R is the radius of the central body.


What does this formula tell us?

1. Orbital speed is directly proportional to the square root of the mass of the central body (larger planet, faster orbit for the same radius).


2. Orbital speed is inversely proportional to the square root of the orbital radius (closer to the central body, faster the orbit). This makes sense: gravity is stronger closer to the Earth, so you need to move faster to avoid falling.

Example 1: Orbital Speed of a Low Earth Orbit Satellite
Let's calculate the orbital speed of a satellite orbiting Earth at an altitude of 400 km (typical for the ISS).
Given:
R_Earth = 6371 km = 6.371 × 10⁶ m
M_Earth = 5.972 × 10²⁴ kg
G = 6.674 × 10⁻¹¹ N m²/kg²
Altitude h = 400 km = 0.4 × 10⁶ m

Step 1: Calculate the orbital radius 'r'.
r = R_Earth + h = 6.371 × 10⁶ m + 0.4 × 10⁶ m = 6.771 × 10⁶ m

Step 2: Use the orbital speed formula.
v_o = √(GM / r)
v_o = √((6.674 × 10⁻¹¹ N m²/kg²) * (5.972 × 10²⁴ kg) / (6.771 × 10⁶ m))
v_o = √(3.98 × 10¹⁴ / 6.771 × 10⁶)
v_o = √(5.877 × 10⁷)
v_o ≈ 7666 m/s or about 7.67 km/s.

That's incredibly fast! To put it in perspective, that's over 27,000 kilometers per hour!

### 4. Time Period (T): How Long Does One Orbit Take?

The time period (T) of a satellite is the time it takes to complete one full revolution around the central body.
For a circular orbit, the satellite travels a distance equal to the circumference of the orbit (2πr) in one time period (T) at a constant speed (v_o).

So, we can use the basic formula: Time = Distance / Speed

T = 2πr / v_o

Now, substitute the expression for v_o that we just derived (v_o = √(GM / r)):

T = 2πr / √(GM / r)

T = 2πr * √(r / GM)

T = 2π * √(r² * r / GM) (Here, we moved 'r' inside the square root by squaring it)

T = 2π√(r³ / GM)

This formula shows us:

1. The time period depends on the mass of the central body and the orbital radius.


2. For a larger orbital radius, the time period is longer. This makes intuitive sense: you're traveling a longer path, and you're moving slower (as v_o decreases with r).

A fascinating consequence of this formula is Kepler's Third Law of Planetary Motion for circular orbits:

If you square both sides of the time period equation:

T² = (2π)² * (r³ / GM)

T² = (4π² / GM) * r³

Since (4π² / GM) is a constant for a given central body, we can write:

T² ∝ r³

This means that the square of the orbital period is directly proportional to the cube of the orbital radius. This law applies to all objects orbiting the same central body (e.g., all planets orbiting the Sun, or all satellites orbiting the Earth).

Example 2: Time Period of the ISS (cont.)
Using the orbital radius r = 6.771 × 10⁶ m and M_Earth = 5.972 × 10²⁴ kg, G = 6.674 × 10⁻¹¹ N m²/kg².

Step 1: Use the time period formula.
T = 2π√(r³ / GM)
T = 2π√((6.771 × 10⁶ m)³ / ((6.674 × 10⁻¹¹ N m²/kg²) * (5.972 × 10²⁴ kg)))
T = 2π√((3.10 × 10²⁰) / (3.98 × 10¹⁴))
T = 2π√(7.788 × 10⁵)
T = 2π * 882.5 seconds
T ≈ 5545 seconds

Step 2: Convert to minutes/hours.
5545 seconds / 60 seconds/minute ≈ 92.4 minutes.
This means the ISS completes one orbit around Earth in about 92.4 minutes! That's why astronauts see so many sunrises and sunsets in a single "day".

### 5. Energy of a Satellite: Staying in the Gravitational Well

Energy is a crucial concept in understanding bound systems like satellites in orbit. A satellite possesses two types of mechanical energy: Kinetic Energy (due to its motion) and Gravitational Potential Energy (due to its position in the gravitational field).

#### a) Kinetic Energy (KE)

The kinetic energy of a satellite of mass 'm' moving with orbital speed 'v_o' is given by:

KE = ½ mv_o²

We already know v_o² = GM/r. Let's substitute this:

KE = ½ m (GM/r) = GMm / (2r)

Notice that KE is always positive, as expected. It's inversely proportional to the orbital radius 'r'. This means as 'r' increases, KE decreases (the satellite moves slower).

#### b) Gravitational Potential Energy (PE)

The gravitational potential energy of a satellite of mass 'm' at a distance 'r' from the center of a central body of mass 'M' is given by:

PE = -GMm / r

Why the negative sign? This negative sign is very important. It signifies that the satellite is in a "bound system" or a "gravitational well." The potential energy is defined to be zero when the satellite is infinitely far away (r = ∞). As the satellite gets closer to the central body, gravity does positive work on it, and its potential energy decreases, becoming negative. A more negative potential energy means it's more tightly bound to the central body.

#### c) Total Mechanical Energy (E_T)

The total mechanical energy of a satellite in orbit is the sum of its kinetic energy and potential energy:

E_T = KE + PE

E_T = (GMm / (2r)) + (-GMm / r)

E_T = (GMm / (2r)) - (2GMm / (2r))

E_T = -GMm / (2r)

This total energy for a satellite in a circular orbit is also negative! This negative total energy reinforces the idea of a bound system. If the total energy were positive, the satellite would escape the gravitational influence of the central body and fly off into space. If it were zero, it would escape to infinity with zero kinetic energy.


Let's look at the fascinating relationships between KE, PE, and E_T:

From our derivations:
* KE = GMm / (2r)
* PE = -GMm / r
* E_T = -GMm / (2r)

Comparing these, we can see:
* KE = -E_T (Kinetic energy is the negative of the total energy).
* PE = 2E_T (Potential energy is twice the total energy).
* PE = -2KE (Potential energy is negative two times the kinetic energy).

These relationships are extremely useful in solving problems related to satellite energy.

Example 3: Energy of the ISS (cont.)
Let's consider a 400 kg satellite (a small part of the ISS for simplicity) orbiting Earth at r = 6.771 × 10⁶ m.
m = 400 kg
M_Earth = 5.972 × 10²⁴ kg
G = 6.674 × 10⁻¹¹ N m²/kg²
r = 6.771 × 10⁶ m

Step 1: Calculate KE.
KE = GMm / (2r)
KE = (6.674 × 10⁻¹¹ * 5.972 × 10²⁴ * 400) / (2 * 6.771 × 10⁶)
KE = (1.594 × 10¹⁷) / (1.3542 × 10⁷)
KE ≈ 1.177 × 10¹⁰ Joules

Step 2: Calculate PE.
PE = -GMm / r
PE = -(6.674 × 10⁻¹¹ * 5.972 × 10²⁴ * 400) / (6.771 × 10⁶)
PE = -(1.594 × 10¹⁷) / (6.771 × 10⁶)
PE ≈ -2.354 × 10¹⁰ Joules

Step 3: Calculate E_T.
E_T = KE + PE
E_T = 1.177 × 10¹⁰ J + (-2.354 × 10¹⁰ J)
E_T ≈ -1.177 × 10¹⁰ Joules

Check relationships:
KE = 1.177 × 10¹⁰ J
-E_T = -(-1.177 × 10¹⁰ J) = 1.177 × 10¹⁰ J (Matches KE!)
PE = -2.354 × 10¹⁰ J
2E_T = 2 * (-1.177 × 10¹⁰ J) = -2.354 × 10¹⁰ J (Matches PE!)
The relations hold true!

### 6. Summary & Key Takeaways

Concept	Formula	Key Dependence	Physical Significance
Orbital Speed (v_o)	v_o = √(GM / r)	M (central body mass), r (orbital radius). Independent of satellite's mass 'm'.	The speed required to maintain a stable circular orbit at radius 'r'. Faster for smaller 'r'.
Time Period (T)	T = 2π√(r³ / GM) or T² ∝ r³ (Kepler's 3rd Law)	M (central body mass), r (orbital radius).	Time for one complete revolution. Longer for larger 'r'.
Kinetic Energy (KE)	KE = GMm / (2r)	G, M, m, r.	Energy due to motion. Always positive. Decreases with increasing 'r'.
Potential Energy (PE)	PE = -GMm / r	G, M, m, r.	Energy due to position in gravitational field. Always negative for bound orbits. More negative for smaller 'r'.
Total Energy (E_T)	E_T = -GMm / (2r)	G, M, m, r.	Sum of KE and PE. Always negative for bound orbits, indicating the satellite is gravitationally "trapped".

CBSE vs. JEE Focus: For both CBSE and JEE, understanding these fundamental derivations and the physical meaning of each quantity is paramount. CBSE might focus more on direct application of formulas and conceptual understanding, while JEE will expect you to manipulate these formulas in more complex scenarios, combine them with other concepts (like energy conservation, escape velocity), and solve multi-step problems. A strong grasp of these basics is your foundation for both!

Keep practicing with different values and scenarios, and you'll master satellite motion in no time!

Welcome, aspiring physicists! Today, we embark on a deep dive into the fascinating world of satellites, specifically exploring their orbital characteristics: orbital speed, time period, and energy. This topic is fundamental to understanding how artificial satellites stay in orbit and how celestial bodies interact, making it crucial for both your conceptual clarity and JEE exam preparation.

We'll start with the basics, assume a perfectly circular orbit for simplicity (though real orbits are often elliptical, the circular orbit is an excellent approximation and forms the foundation), and gradually build up the complexities.

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### 1. The Mechanics of Satellite Motion: A Recap

Before we delve into the specifics, let's quickly recall the fundamental forces at play when a satellite orbits a planet:

1. Newton's Law of Universal Gravitation: This law states that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
For a satellite of mass 'm' orbiting a planet of mass 'M' at a distance 'r' (from the center of the planet), the gravitational force is:

F_g = GMm/r²

Here, 'G' is the universal gravitational constant ($6.67 imes 10^{-11} ext{ Nm}^2/ ext{kg}^2$).

2. Centripetal Force: For an object to move in a circular path, a force directed towards the center of the circle is required. This is the centripetal force, given by:

F_c = mv²/r

Where 'm' is the mass of the object, 'v' is its speed, and 'r' is the radius of the circular path.

In the case of a satellite orbiting a planet, the gravitational force *provides the necessary centripetal force* for the satellite to maintain its circular orbit.

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### 2. Orbital Speed of a Satellite (v₀)

Imagine a satellite moving around the Earth in a stable circular orbit. What speed does it need to maintain this path? This is its orbital speed.

#### Derivation:

For a stable circular orbit, the gravitational force (F_g) must be equal to the centripetal force (F_c):

F_g = F_c

GMm/r² = mv₀²/r

Notice that the mass of the satellite, 'm', cancels out from both sides! This is a crucial insight: the orbital speed required for a given orbit is independent of the mass of the satellite.

Solving for v₀:

v₀² = GM/r

v₀ = √(GM/r)

Where:
* G: Universal Gravitational Constant
* M: Mass of the central body (e.g., Earth)
* r: Radius of the orbit, measured from the center of the central body. If the satellite is at an altitude 'h' above the surface, then r = R + h, where 'R' is the radius of the central body.

#### Key Takeaways for Orbital Speed:

* Independence from Satellite Mass: A small satellite and a large space station, at the same altitude, will have the same orbital speed. This is similar to how all objects fall with the same acceleration (g) in a vacuum, regardless of their mass.
* Dependence on Central Body Mass (M): A more massive central body requires a higher orbital speed for a given radius.
* Dependence on Orbital Radius (r): As the orbital radius 'r' increases (i.e., higher altitude), the orbital speed 'v₀' decreases. This is intuitive: farther objects experience less gravitational pull and thus need less speed to stay in orbit.
* Relation to 'g' (acceleration due to gravity): We know that on the surface of Earth, g = GM/R². At an altitude 'h', the effective 'g' is g' = GM/r² = GM/(R+h)².
We can write v₀ = √(g'r).

#### Example 1: Calculating Orbital Speed

A satellite is orbiting the Earth at an altitude of 300 km above its surface. Calculate its orbital speed.
(Given: Mass of Earth M = $5.97 imes 10^{24}$ kg, Radius of Earth R = $6.37 imes 10^6$ m, G = $6.67 imes 10^{-11} ext{ Nm}^2/ ext{kg}^2$)

Step-by-step Solution:

1. Determine the orbital radius (r):
Altitude h = 300 km = $300 imes 10^3$ m
Radius of Earth R = $6.37 imes 10^6$ m
r = R + h = $(6.37 imes 10^6) + (300 imes 10^3)$ m
r = $(6.37 imes 10^6) + (0.30 imes 10^6)$ m = $6.67 imes 10^6$ m

2. Apply the orbital speed formula:
v₀ = √(GM/r)
v₀ = √[($6.67 imes 10^{-11} ext{ Nm}^2/ ext{kg}^2$) * ($5.97 imes 10^{24}$ kg) / ($6.67 imes 10^6$ m)]
v₀ = √[($5.97 imes 10^{13}$) / ($10^6$)]
v₀ = √($5.97 imes 10^7$)
v₀ ≈ 7726 m/s or 7.73 km/s

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### 3. Time Period of a Satellite (T)

The time period of a satellite is the time it takes to complete one full revolution around its central body.

#### Derivation:

For a circular orbit, the distance covered in one revolution is the circumference of the orbit (2πr). Since distance = speed × time, we have:

T = Distance / Speed = 2πr / v₀

Now, substitute the expression for v₀ = √(GM/r):

T = 2πr / √(GM/r)

T = 2πr * √(r/GM)

T = 2π * √(r³/GM)

To eliminate the square root, we can square both sides:

T² = (2π)² * (r³/GM)

T² = (4π²/GM) * r³

T = 2π√(r³/GM)

This equation is a direct manifestation of Kepler's Third Law of Planetary Motion, which states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit (for a circular orbit, the semi-major axis is simply 'r').

#### Key Takeaways for Time Period:

* Independence from Satellite Mass: Just like orbital speed, the time period is independent of the satellite's mass.
* Dependence on Central Body Mass (M): A more massive central body leads to a shorter time period for a given radius (satellites orbit faster).
* Dependence on Orbital Radius (r): As the orbital radius 'r' increases, the time period 'T' increases significantly (T ∝ r³/²). Satellites at higher altitudes take longer to complete an orbit.
* JEE FOCUS: The term (4π²/GM) is a constant for orbits around a given central body. This means T² ∝ r³. This proportionality is extremely useful for comparing the periods of different satellites orbiting the same central body.

#### Example 2: Calculating Time Period

Using the satellite from Example 1 (orbital radius r = $6.67 imes 10^6$ m), calculate its time period.

Step-by-step Solution:

1. Apply the time period formula:
T = 2π√(r³/GM)
T = 2π√[($6.67 imes 10^6$ m)³ / (($6.67 imes 10^{-11} ext{ Nm}^2/ ext{kg}^2$) * ($5.97 imes 10^{24}$ kg))]
T = 2π√[($2.964 imes 10^{20}$) / ($3.98 imes 10^{14}$)]
T = 2π√($7.447 imes 10^5$)
T = 2π * 863.07 s
T ≈ 5422 seconds

2. Convert to minutes/hours for better understanding:
T ≈ 5422 / 60 minutes ≈ 90.37 minutes
T ≈ 90.37 / 60 hours ≈ 1.51 hours

This means the satellite completes one orbit around Earth in about 90 minutes. This is typical for Low Earth Orbit (LEO) satellites.

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### 4. Energy of a Satellite

The total mechanical energy of a satellite in orbit is the sum of its kinetic energy and gravitational potential energy. Understanding satellite energy is crucial for analyzing orbital transfers, stability, and the energy required for launch.

#### A. Kinetic Energy (K.E.)

The kinetic energy of a satellite of mass 'm' moving with orbital speed 'v₀' is:

K.E. = ½ mv₀²

Substitute v₀² = GM/r (from our orbital speed derivation):

K.E. = ½ m (GM/r) = GMm / 2r

#### B. Gravitational Potential Energy (P.E. or U)

The gravitational potential energy of a satellite of mass 'm' in the gravitational field of a central body of mass 'M' at a distance 'r' is defined as:

P.E. = -GMm / r

The negative sign is important. It indicates that the system is a bound system. Potential energy is conventionally taken as zero at infinite separation (r → ∞). As the satellite moves closer to the central body, it loses potential energy, meaning its potential energy becomes more negative. This negative value reflects the attractive nature of gravity.

#### C. Total Mechanical Energy (E)

The total mechanical energy of the satellite is the sum of its kinetic energy and potential energy:

E = K.E. + P.E.

E = (GMm / 2r) + (-GMm / r)

E = (GMm / 2r) - (2GMm / 2r)

E = -GMm / 2r

#### Key Takeaways for Total Energy:

* Negative Total Energy: The total mechanical energy of a satellite in a stable circular orbit is always negative. This is the hallmark of a bound system. It means that the satellite is gravitationally bound to the central body and requires an input of energy to escape its orbit (i.e., to reach zero or positive total energy).
* Relationship between K.E., P.E., and E:
* K.E. = -E
* P.E. = 2E
* P.E. = -2 K.E.
These relationships are very useful for quick calculations in JEE problems.
* Dependence on Orbital Radius (r): As 'r' increases (higher altitude), 'E' becomes less negative (i.e., increases). This means a satellite at a higher orbit has more energy (is less tightly bound) than one in a lower orbit.
* Conservation of Energy: In the absence of non-conservative forces like air drag or thrust, the total mechanical energy of a satellite remains constant throughout its orbit.

#### D. Binding Energy

The binding energy is the minimum energy required to completely remove the satellite from its orbit and take it to infinity with zero kinetic energy. It is essentially the positive value of the total mechanical energy.

ΔE = GMm (1/R - 1/2r)

This energy is provided by the rocket engines.

#### Example 3: Calculating Satellite Energy

Consider a 100 kg satellite orbiting the Earth at an altitude of 300 km (r = $6.67 imes 10^6$ m). Calculate its kinetic energy, potential energy, and total mechanical energy.
(Given: M = $5.97 imes 10^{24}$ kg, G = $6.67 imes 10^{-11} ext{ Nm}^2/ ext{kg}^2$, m = 100 kg)

Step-by-step Solution:

1. Calculate Kinetic Energy (K.E.):
K.E. = GMm / 2r
K.E. = ($6.67 imes 10^{-11}$) * ($5.97 imes 10^{24}$) * 100 / (2 * $6.67 imes 10^6$)
K.E. = ($3.98 imes 10^{16}$) / ($1.334 imes 10^7$)
K.E. ≈ $2.98 imes 10^9$ Joules

2. Calculate Potential Energy (P.E.):
P.E. = -GMm / r
P.E. = -($6.67 imes 10^{-11}$) * ($5.97 imes 10^{24}$) * 100 / ($6.67 imes 10^6$)
P.E. = -($3.98 imes 10^{16}$) / ($6.67 imes 10^6$)
P.E. ≈ -$5.97 imes 10^9$ Joules

*Cross-check:* P.E. should be approximately -2 * K.E.
-2 * ($2.98 imes 10^9$) = -$5.96 imes 10^9$ J, which matches very well.

3. Calculate Total Mechanical Energy (E):
E = K.E. + P.E.
E = ($2.98 imes 10^9$) + (-$5.97 imes 10^9$)
E = -$2.99 imes 10^9$ Joules

*Cross-check:* E should be -K.E.
-K.E. = -($2.98 imes 10^9$) J, which also matches closely due to rounding.
Also, E should be P.E./2.
P.E./2 = (-$5.97 imes 10^9$) / 2 = -$2.985 imes 10^9$ J. All checks confirm the results.

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### 5. Important Relations and JEE Insights

Let's consolidate some critical relationships and points often tested in JEE.

| Parameter | Formula (Circular Orbit) | Key Dependence | Independence | Remarks |
| :------------------ | :-------------------------------------------- | :--------------------------------------- | :---------------------------- | :----------------------------------------------------------------- |
| Orbital Speed (v₀) | √(GM/r) | M, r (decreases with r) | Satellite mass (m) | Crucial for maintaining orbit, v₀ < v_escape |
| Time Period (T) | 2π√(r³/GM) or T² ∝ r³ | M, r (increases significantly with r) | Satellite mass (m) | Kepler's Third Law, basis for geostationary orbits |
| Kinetic Energy (K.E.) | GMm/2r | M, m, r (decreases with r) | - | Always positive |
| Potential Energy (P.E.) | -GMm/r | M, m, r (becomes less negative with r) | - | Always negative, zero at infinity |
| Total Energy (E) | -GMm/2r | M, m, r (becomes less negative with r) | - | Always negative, indicates a bound system, E = -K.E. = P.E./2 |
| Binding Energy | GMm/2r | M, m, r (decreases with r) | - | Energy required to escape orbit |

* Relationship between Orbital Speed (v₀) and Escape Speed (vₑ):
Escape speed from a distance 'r' is vₑ = √(2GM/r).
Therefore, v₀ = vₑ/√2. This means the orbital speed is √2 times *smaller* than the escape speed at the same radius. If a satellite's speed increases to vₑ, it will escape the gravitational pull.

* Types of Orbits based on Energy:
* E < 0: Bound system (elliptical or circular orbit).
* E = 0: Parabolic trajectory (just escapes).
* E > 0: Hyperbolic trajectory (escapes with residual speed).

* Geostationary Satellites: These are a special case where T = 24 hours (Earth's rotational period), and they orbit in the equatorial plane, appearing stationary from the ground. This requires a specific orbital radius (approximately 42,164 km from Earth's center or 35,786 km above the surface).

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By thoroughly understanding these derivations, relationships, and the physical significance of each term, you will be well-equipped to tackle a wide range of problems related to satellite motion in your JEE examinations and beyond. Keep practicing, and visualize these concepts!

Understanding and recalling the key formulas for satellite motion – orbital speed, time period, and energy – is crucial for both JEE and board exams. Here are some mnemonics and short-cuts to help you remember these complex expressions and their relationships.

### Orbital Speed ($v_o$)

The formula for orbital speed is $v_o = sqrt{frac{GM}{r}}$.

* Mnemonic: "Very Good Man, Root it!"
* Very $
ightarrow v_o$
* Good $
ightarrow G$ (Gravitational constant)
* Man $
ightarrow M$ (Mass of the central body, e.g., Earth)
* Root it $
ightarrow sqrt{dots / r}$ (Radius of the orbit from the center of the central body)

* Short-Cut Tip: Compare it with escape velocity ($v_e = sqrt{frac{2GM}{R}}$). Notice the '2' factor and 'R' (radius of planet) versus 'r' (orbital radius). The orbital speed is generally lower than escape speed at the same radius.

### Time Period (T)

The formula for the time period of a satellite is $T = 2pisqrt{frac{r^3}{GM}}$.

* Mnemonic: "Two Pirates Roam Constantly, Grab Money."
* Two Pirates $
ightarrow 2pi$
* Roam Constantly $
ightarrow sqrt{r^3}$ (Cube of the orbital radius 'r')
* Grab Money $
ightarrow /GM$ (divided by G times M)

* Kepler's Third Law (Short-Cut): For many problems, especially ratio-based ones, remembering Kepler's Third Law is a major short-cut:
* $T^2 propto r^3$ or $frac{T_1^2}{T_2^2} = frac{r_1^3}{r_2^3}$
* Mnemonic: "Time Squared is Radius Cubed" (TS-RC)

### Energy of a Satellite: Kinetic (K), Potential (U), Total (E)

Remembering the sign and magnitude relationships between Kinetic (K), Potential (U), and Total (E) energies is crucial for problem-solving.
* $K = frac{GMm}{2r}$ (always positive)
* $U = -frac{GMm}{r}$ (always negative for attractive forces)
* $E = -frac{GMm}{2r}$ (always negative for bound orbits)

* Mnemonic for Signs:
* Kinetic energy is Positive. (Kids are Positive!)
* Unique and Evil energies are Negative. (U & E are Negative!)

* Mnemonic for Magnitude Relationships: "KUNE Easy 2 Remember!"
* $|U| = 2|K|$ (Potential magnitude is twice Kinetic magnitude)
* $|E| = |K|$ (Total energy magnitude is equal to Kinetic magnitude)
* Combine with signs:
* $E = -K$ (Total energy is negative of Kinetic energy)
* $U = -2K$ (Potential energy is negative twice Kinetic energy)
* $E = U/2$ (Total energy is half of Potential energy)

Energy Type	Formula	Mnemonic/Relationship
Kinetic (K)	$frac{GMm}{2r}$	Positive always
Potential (U)	$-frac{GMm}{r}$	Negative always $U = -2K$
Total (E)	$-frac{GMm}{2r}$	Negative always $E = -K$ $E = U/2$

### JEE/CBSE Exam Tip:
For JEE Advanced, understanding the derivation and the interplay of these formulas is crucial. For JEE Main and CBSE, quick recall using these mnemonics can save valuable time, especially in multiple-choice questions involving relative magnitudes or signs of energies. Always pay attention to the value of 'r' (distance from the center of the Earth) versus 'h' (height above the surface). Remember $r = R_e + h$.

Keep practicing with these mnemonics to make them second nature!

🚀 Quick Tips: Orbital Speed, Time Period, and Energy of Satellite

Understanding the mechanics of satellite motion, particularly orbital speed, time period, and energy, is crucial for both JEE Main and board exams. These concepts are frequently tested and require a firm grasp of their formulas and interrelationships.

Here are some quick tips to master this topic:

1. Orbital Speed (v)

• Formula for Circular Orbit: The orbital speed of a satellite in a circular orbit of radius 'r' around a planet of mass 'M' is given by:
  
  v = √(GM/r)
  
  Where G is the universal gravitational constant.
  


• Independence of Satellite Mass: Note that the orbital speed does not depend on the mass of the satellite (m). It only depends on the mass of the central body (M) and the orbital radius (r).


• Relationship with Escape Velocity: For a satellite orbiting very close to the Earth's surface (r ≈ R), orbital speed (v₀) and escape velocity (vₑ) are related as:
  
  v₀ = vₑ / √2
  


• JEE Tip: Pay attention to whether the question asks for orbital speed *at the surface* or *at a certain height* 'h' (in which case r = R+h).

2. Time Period (T)

• Formula for Time Period: The time period (T) is the time taken for one complete revolution.
  
  T = 2πr / v = 2πr * √(r/GM) = 2π√(r³/GM)
  


• Kepler's Third Law: From the formula, it's clear that T² ∝ r³. This is Kepler's Law of Periods. This proportionality is extremely useful for comparing time periods of satellites at different radii.


• Independence of Satellite Mass: Like orbital speed, the time period also does not depend on the mass of the satellite.


• Geostationary Satellite: A geostationary satellite has a time period of 24 hours (or 86400 seconds) and orbits at an approximate height of 35,800 km above the Earth's surface (r ≈ 42,200 km from Earth's center).

3. Energy of Satellite

• Kinetic Energy (KE): KE = ½ mv² = ½ m (GM/r) = GMm / (2r)


• Potential Energy (PE): The gravitational potential energy is always negative, signifying a bound system.
  
  PE = -GMm / r
  


• Total Mechanical Energy (E): The sum of kinetic and potential energy:
  
  E = KE + PE = (GMm / 2r) + (-GMm / r) = -GMm / (2r)
  


• Key Relationships:
  
  
  
  • E = -KE
  
  
  • PE = 2E
  
  
  • PE = -2KE
  
  
  
  These relationships are vital for quickly solving problems involving energy changes.
  


• Energy for Orbit Transfer: To transfer a satellite from a lower orbit to a higher orbit, energy must be supplied (i.e., total energy must increase, becoming less negative). The energy required to launch a satellite from Earth's surface to a specific orbit is the difference between its total energy in orbit and its initial energy at the surface.


• Total Energy is Negative: A negative total energy indicates that the satellite is gravitationally bound to the central body and requires energy input to escape its gravitational pull.

Stay focused on these core formulas and relationships. Practice problems involving changes in orbital parameters to solidify your understanding!

Intuitive Understanding: Orbital Speed, Time Period, and Energy of a Satellite

Understanding the motion of satellites around a celestial body like Earth requires grasping the fundamental concepts of orbital speed, time period, and energy. These aren't just formulas to memorize; they represent a delicate balance of forces and energies.

1. Orbital Speed: The "Just Right" Velocity

Imagine throwing a stone horizontally. If you throw it too slowly, it quickly falls to the ground. If you throw it faster, it travels further before falling. Now imagine throwing it so fast that as it falls, the Earth's surface curves away beneath it, and it continuously "misses" the ground. This "missing" the ground while continuously falling is what we call orbital motion, and the speed required is the orbital speed.

• 
  The Balance Act: A satellite stays in orbit because of a precise balance between two factors:
  
  
  
  • Gravity: Pulling the satellite towards the Earth.
  
  
  • Inertia (Tendency to Move in a Straight Line): The satellite's forward motion that tries to pull it away from the Earth.
  
  
  
  If the speed is too low, gravity wins, and the satellite crashes. If the speed is too high, inertia wins, and the satellite escapes Earth's gravity. The orbital speed is the "sweet spot" where these two effects perfectly balance, causing the satellite to move in a stable circular or elliptical path.
  


• 
  Why Closer Orbits are Faster: The closer a satellite is to Earth, the stronger Earth's gravitational pull. To counteract this stronger pull and maintain a stable orbit, the satellite needs to travel faster. Think of it like swinging a ball on a string: to keep it from falling, you have to swing it faster if the string is shorter.
  

2. Time Period: The Orbital Lap Time

The time period (T) of a satellite is simply the time it takes for the satellite to complete one full revolution around the central body (e.g., Earth).

• 
  Factors Influencing Time Period:
  
  
  
  • Orbital Speed: A faster satellite will complete its orbit in less time.
  
  
  • Orbital Radius: A larger orbit means a longer path to cover, which will generally lead to a longer time period, even if the speed is similar or slightly different.
  
  
  
  


• 
  Kepler's Insight: Intuitively, Kepler's Third Law (though quantitative) tells us that for stable orbits, satellites in larger orbits have significantly longer time periods. This is because they travel a longer distance and, counter-intuitively perhaps, actually travel *slower* than satellites in lower orbits.
  

3. Energy of a Satellite: Bound or Free?

The total mechanical energy (E) of a satellite in orbit is the sum of its kinetic energy (KE) due to its motion and its gravitational potential energy (PE) due to its position in the gravitational field.

• 
  The Significance of Negative Energy (JEE Concept):
  
  
  
  • For a satellite in a bound orbit (i.e., orbiting Earth and not escaping), its total mechanical energy is always negative.
  
  
  • Intuitively, a negative total energy means the satellite is "trapped" by Earth's gravity. It signifies that you would need to *add* energy to the satellite to make it escape Earth's gravitational pull and reach a state of zero total energy (where it is infinitely far away and at rest).
  
  
  • Think of it like a ball at the bottom of a valley. Its potential energy is negative relative to the top of the surrounding hills. You need to provide energy to get it out of the valley. Similarly, a satellite is in a "gravitational valley."
  
  
  
  


• 
  Changing Orbits:
  
  
  
  • To move a satellite to a higher orbit, you need to *increase* its total energy (make the negative value closer to zero). This usually involves firing rockets to provide a thrust, increasing its kinetic energy.
  
  
  • To bring a satellite to a lower orbit, you need to *decrease* its total energy (make the negative value more negative). This often involves retro-firing rockets to slow it down, causing it to fall to a lower, more negative energy state.
  
  
  
  


• 
  Energy Partition: For a circular orbit, the kinetic energy is positive, and the potential energy is negative and twice the magnitude of the kinetic energy, making the total energy negative. This is a crucial relationship often tested in JEE.
  

By understanding these intuitive concepts, you'll find the mathematical derivations and problem-solving for satellite motion much more accessible.

Real World Applications of Orbital Speed, Time Period, and Energy of Satellites

The principles governing a satellite's orbital speed, time period, and energy are not merely theoretical concepts; they form the bedrock of virtually all space-based technologies that impact our daily lives. Understanding these physics principles is crucial for designing, launching, and maintaining functional satellites.

Here are some key real-world applications:

• 
  Communication Satellites (Geostationary Orbits):
  
  
  
  • Concept: For a satellite to appear stationary from Earth's surface (geostationary), its orbital period must be exactly 24 hours, and it must orbit in the equatorial plane. This requires a very specific orbital speed at a precise altitude (approximately 35,786 km).
  
  
  • Application: This allows communication antennas on Earth to remain fixed, providing uninterrupted TV broadcasting, internet services, and telephone communication across vast regions. The orbital energy required to achieve and maintain this high-altitude orbit is significant and must be carefully managed.
  
  
  
  


• 
  Global Positioning System (GPS) Satellites (Medium Earth Orbits - MEO):
  
  
  
  • Concept: GPS satellites operate at an altitude of about 20,200 km, with an orbital period of approximately 12 hours. Their orbital speed is designed such that they cover the entire Earth in a structured manner.
  
  
  • Application: A constellation of these satellites, each with a precisely calculated trajectory (defined by its speed and period), allows receivers on Earth to triangulate their position, providing accurate navigation for vehicles, smartphones, and aircraft. The orbital energy determines their altitude and thus their coverage pattern.
  
  
  
  


• 
  Earth Observation and Remote Sensing Satellites (Low Earth Orbits - LEO):
  
  
  
  • Concept: Satellites in LEO (altitudes typically between 400-2000 km) have much faster orbital speeds and shorter orbital periods (e.g., 90-120 minutes) compared to MEO or geostationary satellites.
  
  
  • Application: This allows them to pass over specific regions of Earth frequently. They are used for weather forecasting, environmental monitoring, resource management, disaster relief, and military surveillance. Their lower orbital energy makes them cheaper to launch, but they face more atmospheric drag, requiring periodic boosts to maintain orbit. The International Space Station (ISS) is a prime example of a LEO satellite, requiring regular energy input to counteract drag.
  
  
  
  


• 
  Satellite Launch and Deorbiting:
  
  
  
  • Concept: Understanding orbital energy is fundamental to calculating the precise thrust required to lift a satellite from Earth's surface and inject it into a desired orbit with a specific orbital speed and time period.
  
  
  • Application: Similarly, for controlled deorbiting (e.g., for end-of-life satellites to prevent space debris), the satellite's energy must be reduced significantly to allow it to re-enter the atmosphere safely. Incorrect calculations can lead to either failed launches or uncontrolled re-entries.
  
  
  
  

JEE Tip: While CBSE focuses on the derivation and basic understanding, JEE often poses problems that require applying these concepts to scenarios like changes in orbit, energy required for transfers, or comparing orbital parameters of different satellites. A strong grasp of the interrelationships between orbital speed, period, radius, and total energy (kinetic + potential) is crucial.

Understanding complex physics concepts like satellite motion can be made easier by drawing parallels to familiar everyday phenomena. Analogies serve as mental bridges, connecting abstract ideas to concrete experiences. While not perfect representations, they provide intuitive insights into the underlying principles.

1. Orbital Speed: Newton's Cannonball Thought Experiment

• 
  The Analogy: Imagine firing a cannonball horizontally from the top of a very high mountain, high enough to be above the Earth's atmosphere.
  


• 
  Explanation:
  
  
  
  • If you fire it with a low speed, the cannonball follows a parabolic path and falls to Earth (like a normal projectile).
  
  
  • As you increase the firing speed, it travels further before hitting the ground.
  
  
  • At a specific, critical speed (the orbital speed), the cannonball falls towards the Earth, but simultaneously, the Earth's surface curves away beneath it at the same rate. This means it continuously "misses" the Earth and enters a stable orbit.
  
  
  • If fired even faster, it might enter an elliptical orbit or even escape Earth's gravity (if it reaches escape velocity).
  
  
  
  


• 
  Insight: A satellite's orbital speed is precisely the speed required for it to continuously "fall around" the Earth without hitting it or flying off into space. It's a balance between its forward motion (inertia) and the Earth's gravitational pull.

2. Time Period of a Satellite: Running Laps on a Track

• 
  The Analogy: Consider a runner on an athletic track, completing one lap.
  


• 
  Explanation:
  
  
  
  • The time period (T) for a satellite is like the time taken by the runner to complete one full lap.
  
  
  • If the runner maintains a constant speed, the time taken to complete a lap depends on the length of the track (circumference of the orbit). A longer track means a longer time to complete a lap.
  
  
  • Similarly, a satellite in a larger orbit (larger radius) has a longer path to cover. Even though its speed might be slightly lower than in a closer orbit, the increased path length results in a significantly longer time period.
  
  
  
  


• 
  Insight: The time period is directly related to the orbital radius. Larger orbits take longer to complete, which is why geostationary satellites are at a very high altitude (large radius) to have a 24-hour time period.

3. Energy of a Satellite: A Ball in a Gravitational Well

• 
  The Analogy: Imagine a ball at the bottom of a bowl or a "well."
  


• 
  Explanation:
  
  
  
  • 
    The Earth's gravitational field can be thought of as a gravitational well. Objects "fall" into it.
    
  
  
  • 
    For a satellite in orbit, its total mechanical energy (Kinetic Energy + Potential Energy) is negative.
    
  
  
  • 
    Just like a ball at the bottom of a well, its energy is negative relative to the "rim" of the well (which represents infinite distance from Earth, where potential energy is considered zero). As long as the ball's total energy is negative, it remains bound within the well.
    
  
  
  • 
    Similarly, a satellite with negative total energy is bound to the Earth's gravitational field. It will continue to orbit and will not escape into deep space unless its total energy becomes zero or positive.
    
  
  
  • 
    If you give the ball enough energy to reach the rim and go beyond (positive total energy), it escapes the well. This is analogous to a satellite reaching escape velocity.
    
  
  
  
  


• 
  Insight: The negative total energy signifies that the satellite is in a bound state and requires an external input of energy to escape Earth's gravity. For JEE, understanding this negative total energy and its relation to kinetic and potential energy is crucial.

While these analogies simplify complex phenomena, remember that they are aids for conceptual understanding. For competitive exams like JEE, a thorough grasp of the mathematical derivations and formulas for orbital speed, time period, and energy is indispensable.

Prerequisites for Orbital Speed, Time Period, and Energy of Satellite

To effectively understand and solve problems related to the orbital speed, time period, and energy of a satellite, a solid grasp of several foundational physics concepts is crucial. These prerequisites form the building blocks for deriving and applying the relevant equations. Ensure you are comfortable with the following topics:

• Newton's Universal Law of Gravitation:
  
  
  
  • Understanding the formula $F = frac{GMm}{r^2}$ and its vector nature is fundamental. This law quantifies the attractive force between the celestial body (e.g., Earth) and the satellite, which acts as the centripetal force keeping the satellite in orbit.
  
  
  • Relevance: Directly used to determine the gravitational force acting on the satellite, which is the sole force responsible for its orbital motion. (Applicable for both CBSE and JEE).
  
  
  
  


• Circular Motion Dynamics:
  
  
  
  • Concepts like centripetal force ($F_c = frac{mv^2}{r}$ or $F_c = momega^2 r$), centripetal acceleration, angular velocity ($omega$), and linear speed ($v$) are essential. Satellites in stable orbits approximate circular motion.
  
  
  • Relevance: The gravitational force provides the necessary centripetal force for the satellite to move in a circular path. Equating these two forces is the starting point for deriving orbital speed and time period. (Applicable for both CBSE and JEE).
  
  
  
  


• Work, Energy, and Power:
  
  
  
  • A clear understanding of kinetic energy ($KE = frac{1}{2}mv^2$), gravitational potential energy ($U = -frac{GMm}{r}$), and the conservation of mechanical energy is vital.
  
  
  • Relevance: These concepts are directly applied to calculate the total energy of a satellite (sum of kinetic and potential energy) and to understand energy conservation in orbit. (Applicable for both CBSE and JEE).
  
  
  
  


• Gravitational Potential and Potential Energy:
  
  
  
  • Knowing the definition of gravitational potential ($V = -frac{GM}{r}$) and its relationship with potential energy is crucial. Understanding the significance of the negative sign in potential energy is also important.
  
  
  • Relevance: Directly used in the calculation of the potential energy of the satellite, which is a component of its total orbital energy. (More emphasized for JEE derivations and conceptual questions).
  
  
  
  


• Basic Algebra and Equation Manipulation:
  
  
  
  • Proficiency in rearranging equations, solving for unknowns, and handling exponents is necessary to derive and apply the formulas for orbital speed, time period, and energy.
  
  
  • Relevance: Essential for all quantitative problems and derivations. (Crucial for both CBSE and JEE).
  
  
  
  

Mastering these foundational concepts will simplify your understanding of satellite motion and boost your confidence in tackling related problems for both board exams and competitive tests like JEE Main.

Key Takeaways: Orbital Speed, Time Period, and Energy of Satellite

Understanding the dynamics of satellites orbiting a celestial body is fundamental for both JEE Main and board examinations. The orbital speed, time period, and energy of a satellite are interconnected concepts, crucial for solving problems in Gravitation.

• 
  Orbital Speed ($v_o$)
  
  
  
  • 
    Definition: The speed required for a satellite to maintain a stable circular orbit around a central body.
    
  
  
  • 
    Formula: The orbital speed for a satellite of mass 'm' orbiting a central body of mass 'M' at a radius 'r' (from the center of the central body) is given by:
    
    
    $$v_o = sqrt{frac{GM}{r}}$$
    
    
    Where 'G' is the universal gravitational constant.
    
  
  
  • 
    Key Dependence: $v_o$ depends on the mass of the central body (M) and the orbital radius (r).
    
  
  
  • 
    Important Note: Orbital speed is independent of the mass of the satellite (m).
    
  
  
  • 
    Relation to Escape Velocity: At the same radius 'r', the orbital speed $v_o$ is related to the escape velocity $v_e$ by $v_e = sqrt{2} v_o$.
    
  
  
  
  


• 
  Time Period (T)
  
  
  
  • 
    Definition: The time taken by a satellite to complete one full revolution around the central body.
    
  
  
  • 
    Formula: For a circular orbit, it can be derived from $T = frac{2pi r}{v_o}$:
    
    
    $$T = 2pi sqrt{frac{r^3}{GM}}$$
    Or, Kepler's Third Law for circular orbits: $$T^2 = left(frac{4pi^2}{GM}
    ight)r^3$$
    
    
    
  
  
  • 
    Key Dependence: The time period (T) depends on the mass of the central body (M) and the orbital radius (r).
    
  
  
  • 
    Important Note: The time period is independent of the mass of the satellite (m).
    
  
  
  • 
    Geostationary Satellite: A special case where the satellite's time period is 24 hours, making it appear stationary relative to a point on Earth's surface. Its orbital radius is approximately 42,240 km from Earth's center.
    
  
  
  
  


• 
  Energy of Satellite
  
  
  
  • 
    Kinetic Energy (K): For a satellite in a circular orbit, the kinetic energy is:
    
    
    $$K = frac{1}{2}mv_o^2 = frac{GMm}{2r}$$
    
    
    
  
  
  • 
    Gravitational Potential Energy (U): The potential energy of the satellite due to the central body's gravitational field is:
    
    
    $$U = -frac{GMm}{r}$$
    
    
    
  
  
  • 
    Total Mechanical Energy (E): The sum of kinetic and potential energies:
    
    
    $$E = K + U = frac{GMm}{2r} - frac{GMm}{r} = -frac{GMm}{2r}$$
    
    
    
  
  
  • 
    Key Relations: Note the crucial relationships:
    
    
    $$E = -K quad ext{and} quad E = frac{U}{2}$$
    
    
    These relations are very useful for quick problem-solving in JEE.
    
  
  
  • 
    Important Note: The total mechanical energy 'E' is negative. This indicates that the satellite is gravitationally bound to the central body and requires external energy to escape its orbit.
    
  
  
  
  

Master these formulas and their implications to confidently tackle satellite motion problems!

Problem Solving Approach: Orbital Speed, Time Period, and Energy of a Satellite

Solving problems related to satellite motion requires a systematic application of Newton's laws of gravitation and mechanics. Focus on identifying the given parameters and correctly applying the relevant formulas.

1. Understand the Setup and Identify Given Parameters

• Clearly identify the celestial body (e.g., Earth) around which the satellite is orbiting. Its mass is usually denoted as M.


• Identify the mass of the satellite, m.


• Determine the orbital radius, r. This is often given as height h above the surface. Remember, r = R + h, where R is the radius of the central celestial body. If not given, assume it's the radius of Earth (R_E).


• Identify what needs to be calculated: orbital speed (v_o), time period (T), kinetic energy (K), potential energy (U), or total energy (E).

2. Apply Fundamental Principles

The core principle for a satellite in a stable circular orbit is that the gravitational force provides the necessary centripetal force.

Centripetal Force (F_c) = Gravitational Force (F_g)

• F_c = mv_o²/r


• F_g = GMm/r²

Equating these two allows derivation of all other quantities.

3. Key Formulas and Derivations (for quick recall)

1. 
   Orbital Speed (v_o):
   
   
   
   • From mv_o²/r = GMm/r², we get v_o = √(GM/r).
   
   
   • JEE Tip: For a satellite orbiting just above the Earth's surface (r ≈ R), v_o = √(gR), where g is acceleration due to gravity on the surface. Also note v_o = v_e / √2, where v_e is escape velocity.
   
   
   
   


2. 
   Time Period (T):
   
   
   
   • Use the relation T = 2πr / v_o.
   
   
   • Substituting v_o, we get T = 2πr / √(GM/r) = 2π√(r³/GM).
   
   
   • This is Kepler's Third Law (T² ∝ r³).
   
   
   • JEE Tip: Use T² ∝ r³ for comparing periods or radii of different orbits.
   
   
   
   


3. 
   Energy of a Satellite:
   
   
   
   • Potential Energy (U): Due to gravity, U = -GMm/r. Always negative for bound systems.
   
   
   • Kinetic Energy (K): K = ½mv_o². Substitute v_o² = GM/r, so K = GMm/(2r).
   
   
   • Total Energy (E): E = U + K = (-GMm/r) + (GMm/(2r)) = -GMm/(2r).
   
   
   • JEE Tip: Observe the crucial relationships: K = -E and U = 2E. Also, K = -U/2. These relationships are vital for quick problem-solving and conceptual understanding.
   
   
   • Binding Energy: The energy required to remove the satellite from its orbit to infinity is |E| = GMm/(2r).
   
   
   
   

4. Problem-Specific Considerations

• Change in Orbit (JEE Focus): If a satellite moves from one orbit to another, the work done (e.g., by a rocket engine) is equal to the change in its total mechanical energy: Work Done = ΔE = E_final - E_initial.


• Geostationary Satellite: For a geostationary satellite, the time period T = 24 hours. This is a common specific case.

5. Units and Constants

• Always use SI units: meters (m), kilograms (kg), seconds (s).


• Gravitational Constant G = 6.67 × 10⁻¹¹ Nm²/kg².


• Mass of Earth (M_E) ≈ 6 × 10²⁴ kg.


• Radius of Earth (R_E) ≈ 6.4 × 10⁶ m.

6. Review and Verify

After calculation, quickly check if the answer makes physical sense. For instance, the total energy for an orbiting satellite must be negative, and its kinetic energy must be positive.

Remember: Practice deriving these formulas from first principles (Centripetal Force = Gravitational Force). This builds confidence and helps in situations where you might forget a specific formula.

Gravitation is a high-scoring unit in JEE Main, and 'Satellite Motion' is a recurring theme. Mastering orbital speed, time period, and energy of a satellite is crucial for securing marks. Focus on the interrelations and conceptual understanding, not just rote memorization of formulas.

1. Orbital Speed (v_o)

• Definition: The speed required for a satellite to maintain a stable circular orbit around a celestial body.


• Key Formula: Derived by equating gravitational force to centripetal force:
  
  
  $$v_o = sqrt{frac{GM}{r}}$$
  
  
  where G is the universal gravitational constant, M is the mass of the central body (e.g., Earth), and r is the orbital radius (distance from the center of the central body to the satellite, i.e., R + h, where R is radius of central body and h is altitude).
  


• JEE Focus:
  
  
  
  • Notice v_o is independent of the satellite's mass (m).
  
  
  • It decreases as the orbital radius r increases.
  
  
  • Compare with escape velocity: v_e = sqrt{2} v_o for the same orbital radius r.
  
  
  
  

2. Time Period (T)

• Definition: The time taken for a satellite to complete one full revolution around the central body.


• Key Formula:
  
  
  $$T = frac{2pi r}{v_o} = 2pisqrt{frac{r^3}{GM}}$$
  
  
  This formula is a direct application of Kepler's Third Law for circular orbits, stating T^2 propto r^3.
  


• JEE Focus:
  
  
  
  • Like v_o, T is independent of the satellite's mass (m).
  
  
  • T increases significantly with increasing orbital radius r.
  
  
  • Special case: Geostationary satellites have T = 24 ext{ hours} and orbit at a specific altitude.
  
  
  
  

3. Energy of Satellite

Understanding the different forms of energy is crucial for solving problems involving orbital changes.

• Kinetic Energy (K):
  
  
  $$K = frac{1}{2}mv_o^2 = frac{1}{2}mleft(frac{GM}{r}
  ight) = frac{GMm}{2r}$$
  


• Potential Energy (U): Gravitational potential energy (always negative, indicating a bound system).
  
  
  $$U = -frac{GMm}{r}$$
  


• Total Mechanical Energy (E):
  
  
  $$E = K + U = frac{GMm}{2r} - frac{GMm}{r} = -frac{GMm}{2r}$$
  


• Binding Energy (B.E.): The energy required to remove the satellite from its orbit to infinity (where U=0, K=0). It is the positive value of the total energy.
  
  
  $$ ext{B.E.} = -E = frac{GMm}{2r}$$
  

JEE Critical Relationships:

• $K = -E$


• $U = 2E$


• $U = -2K$

Important Note: The negative sign for U and E indicates that the satellite is gravitationally bound to the central body. Positive total energy means the object is unbounded (e.g., moving away, like in the case of escape velocity where E=0).

4. JEE Problem-Solving Strategies

• Conceptual Questions: Often involve understanding how v_o, T, K, U, E change if the orbital radius or central body's mass changes.


• Energy Conservation: For problems involving changes in orbit (e.g., a satellite boosted to a higher orbit), apply the work-energy theorem or conservation of mechanical energy for the entire system (satellite + planet) if only gravitational forces are doing work.
  
  
  $$Delta E = E_{final} - E_{initial} = W_{external}$$
  where W_{external} is work done by external forces (like rocket thrust).
  


• Relative Changes: Be prepared for questions asking for percentage changes or ratios when parameters are altered. For example, if r doubles, how does v_o change? (v_o propto 1/sqrt{r}, so it decreases by sqrt{2} times).

Focus on understanding these relationships thoroughly. A strong grasp of these concepts will enable you to tackle a wide range of problems in JEE Main.