Motion of Satellite

8 MCQs4 revision cards9-step worked example

Source: NCERT GravitationPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

A satellite in circular orbit stays up not because gravity is absent, but because gravity provides the exact centripetal force needed. Confuse "weightlessness inside" with "no gravity at orbit altitude" and you hand NEET a free negative mark.

Orbital velocity. For a satellite of mass m at altitude h above Earth (mass M, radius R), set gravitational pull equal to centripetal force: GMm/(R+h)² = mv²/(R+h). Cancel m and one (R+h) factor to get v = √[GM/(R+h)] (NCERT Class 11 Physics Chapter 7, page 12). Near Earth's surface (h ≈ 0), v₀ = √(gR) ≈ 7.9 km/s.

Time period. Circumference divided by speed: T = 2π(R+h)/v = 2π(R+h)^(3/2)/√(GM). This is Kepler's third law applied to a circular orbit. For a near-surface orbit, T ≈ 84.4 minutes.

Energetics — the signature NEET trap. A bound circular orbit has KE = GMm/[2(R+h)], PE = −GMm/(R+h), and total energy E = −GMm/[2(R+h)]. The relationships: E = −KE = PE/2. A common error is computing only the PE change when asked for the energy to launch a satellite, forgetting that the satellite must also acquire orbital kinetic energy. The energy input equals ΔE = E_orbit − E_surface, not just ΔPE.

Relation to escape velocity. Orbital speed and escape speed at the same radius satisfy v_e = √2 · v_orbital. If orbital speed is boosted by a factor of √2, the satellite escapes.

Watch out: "weightless" astronauts still experience ~90% of surface g at ISS altitude (~400 km). They float because they and the station share the same free-fall orbit — not because gravity vanished.

Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

What is the orbital velocity of a satellite orbiting very close to Earth's surface? (Take g = 9.8 m/s², R = 6.4 × 10⁶ m)

MCQ 2Easy RecallPractice

The time period of a satellite in a circular orbit close to Earth's surface is approximately:

MCQ 3Easy RecallPractice

The ratio of escape velocity to orbital velocity at the same point near Earth's surface is:

MCQ 4Direct ApplicationPractice

A satellite is in a circular orbit of radius r around Earth. Its total mechanical energy is E. If the satellite is moved to a circular orbit of radius 2r, its new total energy is:

MCQ 5Direct ApplicationPractice

For a satellite in a stable circular orbit, which relation is correct?

MCQ 6Direct ApplicationPractice

The energy required to move a satellite of mass m from a circular orbit of radius 2R to 3R (where R is Earth's radius, M is Earth's mass) is:

MCQ 7Concept TrapPractice

An astronaut inside an orbiting space station feels weightless because:

MCQ 8CalculationPractice

A satellite of mass m is launched from Earth's surface (radius R, mass M) into a circular orbit at altitude h = R above the surface. The minimum energy that must be supplied is:

Quick recall before you leave

Worked Example

Pattern: Energy required to launch satellite to altitude — compute change in total mechanical energy (pattern observed in NEET 2024).

1
Given
A satellite of mass m = 200 kg is to be placed in a circular orbit at altitude h = R above Earth's surface. Earth's mass M = 6.0 × 10²⁴ kg, radius R = 6.4 × 10⁶ m, G = 6.67 × 10⁻¹¹ N·m²/kg².
2
Required
Minimum energy that must be supplied to place the satellite in the orbit (starting from rest on the surface).
3
Concept
The satellite starts at the surface with zero KE and PE = −GMm/R. In orbit at radius r = R + h = 2R, its total energy is E = −GMm/(2 × 2R) = −GMm/(4R). The energy input equals ΔE = E_final − E_initial. This accounts for both the gravitational PE raise AND the orbital KE the satellite must acquire — forgetting the KE component is a common error in NEET.
4
Formula
ΔE = E_orbit − E_surface = [−GMm/(4R)] − [−GMm/R] = 3GMm/(4R)
5
Substitution
ΔE = 3 × (6.67 × 10⁻¹¹) × (6.0 × 10²⁴) × (200) / (4 × 6.4 × 10⁶)
6
Calculation
Numerator: 3 × 6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × 200 = 3 × 6.67 × 6.0 × 200 × 10⁻¹¹⁺²⁴ = 3 × 8004 × 10¹³ = 24012 × 10¹³ = 2.4012 × 10¹⁷ Denominator: 4 × 6.4 × 10⁶ = 25.6 × 10⁶ = 2.56 × 10⁷ ΔE = 2.4012 × 10¹⁷ / 2.56 × 10⁷ = 9.38 × 10⁹ J ≈ 9.4 × 10⁹ J Note on exact values: The factors 3, 4, and 200 are exact integers (counting/defined values) and do not limit significant figures. The result is reported to 2 significant figures, governed by the given constants G and M.
7
Final answer
ΔE ≈ 9.4 × 10⁹ J (approximately 9.4 GJ)
8
Common trap
Computing only the PE change: ΔPE = −GMm/(2R) + GMm/R = GMm/(2R) ≈ 6.3 × 10⁹ J. This underestimates the answer because it ignores that the satellite in orbit has KE = GMm/(4R) ≈ 3.1 × 10⁹ J that must also be supplied. The correct answer (3GMm/4R) is 50% larger than the PE-only estimate (GMm/2R).
9
Similar NEET-style question
A rocket launches a 500 kg satellite from Earth's surface into a circular orbit at altitude h = 2R. Find the minimum energy required. (Answer: 5GMm/(6R), computed as E_orbit − E_surface = −GMm/(6R) − (−GMm/R) = 5GMm/(6R).) ---

Before solving, remember these

Formula

Orbital velocity of satellite

For a satellite in circular orbit at altitude h: v = √(G M / (R + h)). At Earth's surface (h ≈ 0): v_orb ≈ √(g R) ≈ 7.9 km/s. Note v_orb = v_e / √2.

-- NCERT Class 11 Physics, Ch. 7, p. 12

Formulas

8 formulas — click to collapse

Escape velocity from a body's surface

v_{e} = \frac{2 GM}{R} = 2 g R

Minimum speed for an object to escape gravity to infinity from radius R. Earth: ~11.2 km/s.

Symbol	Quantity	SI Unit
v_e	escape velocity	m/s
M	planet mass	kg
R	planet radius	m
g	surface gravity	m/s^2

Valid when

Launched from surface
No air drag
Body treated as point/sphere

Gravitational potential energy (point masses)

U = - \frac{GM m}{r}

PE of two-body system; negative because gravity is attractive (work to separate them is positive).

Symbol	Quantity	SI Unit
U	grav PE	J
M, m	two masses	kg
r	separation	m

Valid when

Reference U=0 at r=infinity
Point or spherical masses

g variation with altitude

g_{h} = g (\frac{R}{R + h})^{2} \approx g (1 - 2 h / R)

Gravitational acceleration decreases with altitude above Earth's surface.

Symbol	Quantity	SI Unit
g_h	g at height h	m/s^2
g	surface g	m/s^2
R	Earth radius	m
h	altitude	m

Valid when

Static observer at altitude
Earth treated as uniform sphere

g variation with depth

g_{d} = g (1 - d / R)

Inside Earth (uniform density), g decreases linearly with depth, vanishing at centre.

Symbol	Quantity	SI Unit
g_d	g at depth	m/s^2
g	surface g	m/s^2
R	Earth radius	m
d	depth	m

Valid when

Earth treated as uniform density sphere

Kepler's third law

T^{2} = \frac{4 π ^{2}}{GM} a^{3}

Square of orbital period proportional to cube of semi-major axis. Holds for elliptic orbits about a central mass.

Symbol	Quantity	SI Unit
T	orbital period	s
a	semi-major axis	m
M	central mass	kg

Valid when

Two-body system with central mass M >> orbiting mass
Bound orbit

Orbital velocity for circular orbit

v = \frac{GM}{R + h}

Speed of circular orbit at altitude h above body of mass M, radius R.

Symbol	Quantity	SI Unit
v	orbital speed	m/s
M	central mass	kg
R+h	orbit radius	m

Valid when

Circular orbit
M >> orbiting mass

Satellite total mechanical energy

E = - \frac{GM m}{2 ( R + h )}

Total energy = KE + PE = -KE (virial). Always negative for bound orbit; E -> 0 at infinity.

Symbol	Quantity	SI Unit
E	total energy	J
M, m	central mass and satellite mass	kg
R+h	orbit radius	m

Valid when

Circular orbit
Bound (E < 0)

Newton's law of gravitation

F = \frac{G m _{1} m _{2}}{r ^{2}}

Attractive force between any two masses. Inverse-square central force.

Symbol	Quantity	SI Unit
F	force	N
G	grav constant = 6.674e-11	N*m^2/kg^2
m1, m2	masses	kg
r	centre-to-centre distance	m

Valid when

Point masses or spherically symmetric distributions
r > sum of body radii (else use shell theorem)

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

4 items — click to collapse

Trap

g at altitude: linear vs inverse-square

Category: Similar Terms

Student treats g(h) as linear in h. Actual: g(R/(R+h))² (inverse-square).

When it triggers

Question asks for g at significant altitude (e.g. R/2 above surface).

How to avoid

Use g_h = g (R/(R+h))². Linear approximation g(1 - 2h/R) only valid for h << R.

Trap

Kepler's third law: linear vs 3/2 power

Category: Similar Terms

Student treats T proportional to a (linear) instead of a^(3/2).

When it triggers

Question gives change in semi-major axis and asks for new period.

How to avoid

T² ∝ a³, so T ∝ a^(3/2). Doubling a multiplies T by 2^(3/2) ≈ 2.83.

Mistake

Treats g(h) as linear in h for significant heights.

Root cause: formula misuse

Correction

Use g_h = g(R/(R+h))² (inverse-square). Linear approximation g(1-2h/R) is only valid for h << R. For h = R/2, the exact formula gives g_h = (2/3)² g = 4g/9, not g(1-1) = 0.

Mistake

Treats T proportional to a (linear) instead of a^(3/2) in Kepler's third law.

Root cause: formula misuse

Correction

T² ∝ a³, so T ∝ a^(3/2). Doubling a multiplies T by 2^(3/2) ≈ 2.83. Quadrupling a → 8× T.

Past Year Questions

8 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys. — click to collapse

NEET 2025

The radius of Martian orbit around the Sun is about 4 times the radius of the orbit of Mercury. The Martian year is 687 Earth days. Then which of the following is the length of 1 year on Mercury?

1124 earth days

288 earth days

3225 earth days

4172 earth days

NTA Answer: Option 2(final)

NEET 2025

A body weighs 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is :

136 N

216 N

327 N

432 N

NTA Answer: Option 3(final)

NEET 2024Revised key

1 The mass of a planet is th that of the earth and its diameter is half that of the earth. The acceleration due 10 to gravity on that planet is:

119.6 m s–2

29.8 m s–2

34.9 m s–2

43.92 m s–2

NTA Answer: Option 4(revised_final)

NEET 2024Revised key

The minimum energy required to launch a satellite of mass m from the surface of earth of mass M and radius R in a circular orbit at an altitude of 2R from the surface of the earth is: 5GmM

16R 2GmM

23R GmM

32R GmM

43R

NTA Answer: Option 1(revised_final)

NEET 2023

Two bodies of mass m and 9m are placed at a distance R. The gravitational potential on the line joining the bodies where the gravitational field equals zero, will be (G = gravitational constant) 12Gm 16Gm

1−

2− R R 20Gm 8Gm

3−

4− R R

NTA Answer: Option 2(final)

NEET 2023

A satellite is orbiting just above the surface of the earth with period T. If d is the density of the earth and G is 3π the universal constant of gravitation, the quantity represents Gd

1T2

2T3

NTA Answer: Option 1(final)

NEET 2022

A body of mass 60 g experiences a gravitational force of 3.0 N, when placed at a particular point. The magnitude of the gravitational field intensity at that point is

1180 N/kg

20.05 N/kg

350 N/kg

420 N/kg

NTA Answer: Option 3(final)

NEET 2021

The escape velocity from the Earth's surface is v. The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is

14v

32v

43v

NTA Answer: Option 1(final)

How NEET usually asks this

4 recurring patterns from past papers — click to collapse

Compare escape velocities for planets with different M, R. v_e ∝ sqrt(M/R). Common shape: 'planet has 4× radius and 2× density of Earth — find v_e'.

Direct ApplicationEasy

Common distractors

forgets density not mass

Confuses planet mass with density given radius scaling

Body weighs W on surface; find weight at height h above surface. Use g_h = g(R/(R+h))^2.

Direct ApplicationEasy

Common distractors

uses linear decrease with h

Treats g as linear in h instead of inverse-square

Given period T and semi-major axis a of one planet, find for another with different a. T^2 ∝ a^3.

Direct ApplicationEasy

Common distractors

uses linear relation

Treats T proportional to a not a^(3/2)

Energy required to launch satellite to altitude. Compute change in total mechanical energy.

Multi StepMedium

Common distractors

forgets orbital KE component

Computes only PE change, ignoring orbital KE

Sources

NCERT refs: Class 11 Physics Chapter 7, p.12

Test yourself on this topic with real past-paper questions:

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Keplers Laws Orbital Velocity

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Motion of Satellite

Lesson

Practice MCQs

Quick recall before you leave

v = √[GM/(R+h)]

E = −GMm/[2(R+h)]

T = 2π(R+h)^(3/2)/√(GM)

v_e = √2 · v_orbital

Worked Example

Before solving, remember these

Formulas

Valid when

Valid when

Valid when

Valid when

Valid when

Valid when

Valid when

Valid when

Exam Traps & Common Mistakes

When it triggers

How to avoid

When it triggers

How to avoid

Correction

Correction

Past Year Questions

How NEET usually asks this

Common distractors

Common distractors

Common distractors

Common distractors

Sources

Free NEET study resources