Satellite Time Period

8 MCQs2 revision cards9-step worked example

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

Lesson

Time Period of a Satellite

The trap that costs marks here: treating the satellite's period T as linearly proportional to orbital radius. It is not. The correct relationship is T ∝ r^(3/2), and confusing the exponent turns a 45-second question into a lost mark.

A satellite in circular orbit at radius r from Earth's centre has its gravitational pull supplying the centripetal force. Setting GMm/r² = mv²/r gives the orbital velocity v = √(GM/r). The circumference of the orbit is 2πr, so the time period is:

T = 2πr / v = 2πr / √(GM/r) = 2π√(r³/GM)

Squaring: T² = (4π²/GM) r³

This is Kepler's third law applied to a circular orbit (NCERT Class 11 Physics Chapter 7, page 3). For a satellite at altitude h above Earth's surface, r = R + h, giving T = 2π√((R+h)³ / GM).

Key points:

T depends on r^(3/2), not r. Doubling the orbital radius multiplies the period by 2^(3/2) = 2√2 ≈ 2.83 — not 2.
T is independent of satellite mass — only the central body's mass M matters.
For a near-surface satellite (h ≈ 0), T ≈ 2π√(R³/GM) = 2π√(R/g) ≈ 84.6 minutes for Earth.
The formula connects directly to geostationary orbit design: set T = 24 hours and solve for r.

Watch out: When a question says "orbital radius doubles," reach for the 3/2 power, not the linear scaling. The linear-scaling distractor appears repeatedly in NEET options.

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 time period of a satellite orbiting Earth depends on which of the following?

MCQ 2Easy RecallPractice

What is the approximate time period of a satellite orbiting very close to Earth's surface? (Take R = 6400 km, g = 9.8 m/s²)

MCQ 3Easy RecallPractice

In the formula T² = (4π²/GM)r³ for a satellite's time period, the quantity 4π²/GM is:

MCQ 4Direct ApplicationPractice

A satellite orbits Earth with period T at orbital radius r. If another satellite orbits at radius 4r, what is its period?

MCQ 5Direct ApplicationPractice

Two satellites A and B orbit the same planet. Satellite A has an orbital radius 9 times that of satellite B. The ratio T_A / T_B is:

MCQ 6Direct ApplicationPractice

A geostationary satellite has a period of 24 hours. A satellite in a circular orbit at half the geostationary orbital radius would have a period of:

MCQ 7CalculationPractice

A planet has twice the mass and twice the radius of Earth. What is the ratio of the time period of a near-surface satellite on this planet to that on Earth?

MCQ 8CalculationPractice

A satellite orbits Earth at altitude h = R (where R is Earth's radius) with period T₁. Another satellite orbits at altitude h = 3R. What is the ratio T₂/T₁?

Quick recall before you leave

Worked Example

1
Given
- T_A = 8 years - r_B = 4 r_A
2
Required
Find T_B.
3
Concept
Kepler's third law relates the time period of orbiting bodies to their orbital radii when orbiting the same central mass: T² ∝ r³.
4
Formula
T₂/T₁ = (r₂/r₁)^(3/2) Source: NCERT Class 11 Physics Chapter 7, page 3.
5
Substitution
T_B / T_A = (4 r_A / r_A)^(3/2) = 4^(3/2)
6
Calculation
4^(3/2) = (4^1)(4^(1/2)) = 4 × 2 = 8 Note: The multiplier 4 (the radius ratio) is an exact integer from the problem statement — it does not limit significant figures. T_B = 8 × 8 years = 64 years
7
Final answer
**T_B = 64 years** The quantity 8 (years) is given exactly in the problem, and the ratio 4 is an exact integer, so the answer is exact: 64 years. No significant-figure truncation applies.
8
Common trap
A common mistake is to assume T ∝ r (linear), giving T_B = 4 × 8 = 32 years. This is wrong because Kepler's third law states T² ∝ r³ (i.e. T ∝ r^(3/2)), not T ∝ r. The linear-scaling distractor (32 years) is a high-frequency wrong option in NEET.
9
Similar NEET-style question
If a satellite orbiting Earth has period T at orbital radius r, what is the period of a satellite at orbital radius 9r around the same body? Answer: T_new = T × 9^(3/2) = 27T. ---

Before solving, remember these

Formula

Time period of satellite

T = 2π √((R+h)³ / (G M)) = 2π / ω. For low Earth orbit, T ≈ 84 min. Geostationary orbit: T = 24 h, altitude ≈ 35,786 km.

-- NCERT Class 11 Physics, Ch. 7, p. 13

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.3 | Class 11 Physics Chapter 7, p.13

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Satellite Time Period

Lesson

Practice MCQs

Quick recall before you leave

Satellite time period (circular orbit)

Near-surface satellite period

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