Orbital Velocity

8 MCQs3 revision cards9-step worked example

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

Lesson

Orbital velocity — the speed that keeps a satellite falling around Earth, not into it.

The high-frequency trap in orbital velocity problems: forgetting to include the orbital kinetic energy when computing the energy needed to place a satellite in orbit. Students compute only the gravitational PE change from surface to orbit altitude, losing the KE component — and losing 4 marks.

Core derivation. For a satellite of mass m in circular orbit at radius r = R + h around Earth (mass M), gravity provides the centripetal force:

GMm/r² = mv²/r → v = √(GM/r) = √(GM/(R + h))

This is the orbital velocity (NCERT Class 11 Physics, Chapter 8, page 12). At the surface (h = 0), v₀ = √(GM/R) = √(gR) ≈ 7.9 km/s for Earth.

Key relationships:

Orbital velocity decreases as orbit radius increases: v ∝ 1/√r.
Orbital velocity is independent of the satellite's mass — a 500 kg satellite and a 5000 kg satellite at the same altitude orbit at the same speed.
Relation to escape velocity: v_e = √2 · v_orbital (at the same radius).

Energy in orbit. The total mechanical energy of a satellite in circular orbit is E = −GMm/[2(R + h)]. This equals −KE. The negative sign confirms the orbit is bound. The kinetic energy is KE = GMm/[2(R + h)], and PE = −GMm/(R + h), so PE = 2E and KE = −E.

Watch-out for NEET: When a problem asks "energy required to launch a satellite to orbit at height h," you must compute ΔE = E_orbit − E_surface, which includes both the PE change AND the orbital KE the satellite must acquire. Omitting the KE component is a common error that produces a wrong distractor.

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

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

MCQ 2Easy RecallPractice

The orbital velocity of a satellite depends on:

MCQ 3Easy RecallPractice

For a satellite in a stable circular orbit, the relationship between orbital velocity (v₀) and escape velocity (vₑ) at the same orbital radius is:

MCQ 4Direct ApplicationPractice

A satellite orbits Earth in a circular orbit of radius 2R, where R is Earth's radius. If the orbital velocity near the surface is v₀, the orbital velocity at radius 2R is:

MCQ 5Direct ApplicationPractice

Two satellites of masses m and 4m orbit Earth at the same altitude h. The ratio of their orbital velocities is:

MCQ 6Direct ApplicationPractice

If the radius of Earth were doubled while its mass remained unchanged, the orbital velocity of a near-surface satellite would become:

MCQ 7CalculationPractice

A satellite of mass m is in a circular orbit at height h = R above Earth's surface (R = Earth's radius, M = Earth's mass). The total energy required to move this satellite from the surface of Earth and place it in this orbit is:

MCQ 8CalculationPractice

A satellite is orbiting Earth in a circular orbit of radius r. If its orbital velocity is increased by a factor of √2 (while at radius r), the satellite will:

Quick recall before you leave

Worked Example

1
Given
- m = 200 kg (satellite mass) - h = R = 6.4 × 10⁶ m (orbit altitude equals Earth's radius) - R = 6.4 × 10⁶ m (Earth's radius) - M = 6.0 × 10²⁴ kg (Earth's mass) - G = 6.67 × 10⁻¹¹ N·m²/kg²
2
Required
Minimum energy to move the satellite from rest on the surface to a circular orbit at height h = R (orbit radius = 2R).
3
Concept
The energy required equals the difference in total mechanical energy between the final orbit and the initial surface position. On the surface, the satellite is at rest, so its energy is purely gravitational PE. In orbit, the total energy includes both KE (from orbital motion) and PE.
4
Formulas
- Surface energy: E_surface = −GMm/R (PE only; KE = 0 at rest) - Orbital energy at radius r = 2R: E_orbit = −GMm/(2 · 2R) = −GMm/(4R) - Energy required: ΔE = E_orbit − E_surface
5
Substitution
E_surface = −GMm/R = −(6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × 200) / (6.4 × 10⁶) E_orbit = −GMm/(4R) = −(6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × 200) / (4 × 6.4 × 10⁶)
6
Calculation
First compute GMm = 6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × 200 = 6.67 × 6.0 × 200 × 10⁻¹¹⁺²⁴ = 8.004 × 10¹⁵ J·m Note: The factor 200 (satellite mass) and the 4 in the denominator (from 2R orbit radius substitution) are exact counting/defined integers and do not limit significant figures. The final answer is governed by the 3-significant-figure precision of G and M. E_surface = −8.004 × 10¹⁵ / (6.4 × 10⁶) = −1.251 × 10⁹ J E_orbit = −8.004 × 10¹⁵ / (2.56 × 10⁷) = −3.127 × 10⁸ J ΔE = E_orbit − E_surface = (−3.127 × 10⁸) − (−1.251 × 10⁹) = 9.38 × 10⁸ J
7
Final answer
ΔE ≈ 9.38 × 10⁸ J ≈ 938 MJ Equivalently, in terms of symbols: ΔE = −GMm/(4R) + GMm/R = 3GMm/(4R).
8
Common trap
Computing only the PE change: ΔPE = −GMm/(2R) − (−GMm/R) = GMm/(2R). This gives ~6.25 × 10⁸ J — significantly less than the correct answer because it ignores the kinetic energy the satellite must acquire to maintain orbital velocity at radius 2R. The orbital KE = GMm/(4R) must be added.
9
Similar NEET-style question
A satellite of mass m is launched from Earth's surface to a circular orbit at height h = 2R. Express the minimum energy required in terms of G, M, m, and R. (Answer: 5GMm/(6R) — same method, different orbit radius.) ---

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 8, p.12

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

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Motion of Satellite Satellite Energy

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Orbital Velocity

Lesson

Practice MCQs

Quick recall before you leave

Orbital velocity (circular orbit)

Satellite total mechanical energy

Gravitational potential energy

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