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. 139Satellite Time Period
8 MCQs5 revision cards9-step worked example
Source: NCERT GravitationPYQ coverage: NEET 2023Official key: NTA-verifiedLast updated: 25 Sep 2026
Satellite Time Period, explained for NEET
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 129). 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.
Can you answer these Satellite Time Period 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?
Show answer and why every option is right or wrong
Answer: C. The satellite's time period T = 2π√(r³/GM) depends on the orbital radius r and the central body's mass M only. The satellite's mass cancels out during derivation (NCERT Class 11 Physics Chapter 7, page 139).
Why A is wrong: A is wrong because the satellite's mass cancels when equating gravitational force to centripetal force (GMm/r² = mv²/r). T is independent of m.
Why B is wrong: B is wrong because although gravitational force involves both masses, the satellite's mass cancels in the expression for orbital velocity and period.
Why D is wrong: D is wrong because while orbital radius does affect T, the satellite's mass does not. T depends on r and M (central body mass) only.
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)
Show answer and why every option is right or wrong
Answer: B. For a near-surface satellite, T = 2π√(R/g) = 2π√(6.4 × 10⁶ / 9.8) ≈ 2π × 808 s ≈ 5075 s ≈ 84.6 minutes (NCERT Class 11 Physics Chapter 7, page 139).
Why A is wrong: A is wrong. Computing T = 2π√(R/g) gives approximately 5075 s ≈ 84.6 min, not 60 min. The 60 min figure has no basis in the formula.
Why C is wrong: C is wrong. 120 minutes corresponds to roughly 7200 s, which would require an orbit radius significantly larger than Earth's radius.
Why D is wrong: D is wrong. 24 hours is the period of a geostationary satellite at ~36,000 km altitude, not a near-surface satellite.
MCQ 3Easy RecallPractice
In the formula T² = (4π²/GM)r³ for a satellite's time period, the quantity 4π²/GM is:
Show answer and why every option is right or wrong
Answer: A. 4π²/GM contains only the universal constant G and the central body's mass M. It is the same for every satellite orbiting that body, regardless of orbital radius or satellite mass (NCERT Class 11 Physics Chapter 7, page 129).
Why B is wrong: B is wrong because 4π²/GM contains no satellite mass term — satellite mass cancels in the derivation.
Why C is wrong: C is wrong because 4π²/GM is independent of r. The orbital radius appears separately as r³ on the other side of the equation.
Why D is wrong: D is wrong. The dimensions of 4π²/GM are s²/m³ (time²/length³), which is not dimensionless.
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?
Show answer and why every option is right or wrong
Answer: C. By Kepler's third law, T ∝ r^(3/2). New period = T × (4r/r)^(3/2) = T × 4^(3/2) = T × 8. (NCERT Class 11 Physics Chapter 7, page 129.)
Why A is wrong: A is wrong. This answer assumes T ∝ r (linear scaling). The correct relation is T ∝ r^(3/2), so 4^(3/2) = 8, not 4. (Trap: treating period as linearly proportional to orbital radius.)
Why B is wrong: B is wrong. This would require T ∝ r^(1/2), which is not the correct exponent. Kepler's third law gives T ∝ r^(3/2).
Why D is wrong: D is wrong. This assumes T ∝ r² (square relation), giving 4² = 16. The actual power law is T² ∝ r³, i.e. T ∝ r^(3/2).
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:
Show answer and why every option is right or wrong
Answer: B. T ∝ r^(3/2), so T_A/T_B = (9)^(3/2) = (9^1)(9^(1/2)) = 9 × 3 = 27. (NCERT Class 11 Physics Chapter 7, page 129.)
Why A is wrong: A is wrong. This assumes T ∝ r (linear scaling), giving 9¹ = 9. The correct power is 3/2, not 1. (Trap: linear vs 3/2 power.)
Why C is wrong: C is wrong. This assumes T ∝ r^(1/2), giving 9^(1/2) = 3. The correct exponent from Kepler's third law is 3/2.
Why D is wrong: D is wrong. This assumes T ∝ r² (square relation), giving 9² = 81. Kepler's third law gives T² ∝ r³, hence T ∝ r^(3/2).
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:
Show answer and why every option is right or wrong
Answer: D. T ∝ r^(3/2). New period = 24 × (1/2)^(3/2) = 24 × 1/(2√2) = 24/(2√2) = 12/√2 = 6√2 ≈ 8.49 hours. (NCERT Class 11 Physics Chapter 7, page 129.)
Why A is wrong: A is wrong. 12 hours assumes T ∝ r (linear halving gives half the period). The correct relation T ∝ r^(3/2) yields 24/2^(3/2) = 6√2 hours. (Trap: linear vs 3/2 power.)
Why B is wrong: B is wrong. 3 hours would require T ∝ r³ (halving r divides T by 8). Kepler's law gives T² ∝ r³, i.e. T ∝ r^(3/2).
Why C is wrong: C is wrong. 6 hours would require T ∝ r² (so halving r divides T by 4). The correct exponent is 3/2.
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?
Show answer and why every option is right or wrong
Answer: D. For a near-surface satellite, T = 2π√(R³/GM). Ratio: T_planet/T_Earth = √(R_p³/M_p) / √(R_E³/M_E) = √((2R_E)³/(2M_E)) / √(R_E³/M_E) = √(8R_E³/(2M_E)) × √(M_E/R_E³) = √(4) = 2. (NCERT Class 11 Physics Chapter 7, page 139.)
Why A is wrong: A is wrong. This results from incorrectly computing (2R)³/(2M) as 2R³/M (forgetting to cube the radius factor). The correct calculation: (2R)³ = 8R³, so (8R³)/(2M) = 4R³/M, giving ratio √4 = 2.
Why B is wrong: B is wrong. Doubling both M and R does not leave T unchanged. Since T depends on R³/M, doubling both gives (8R³)/(2M) = 4(R³/M), not 1×(R³/M).
Why C is wrong: C is wrong. 2√2 = √8 comes from cubing the radius factor but forgetting that the mass has doubled too: √((2R)³/M) / √(R³/M) = √8. Dividing by 2M brings it back to √4 = 2.
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?
Show answer and why every option is right or wrong
Answer: A. Orbital radius: r₁ = R + R = 2R; r₂ = R + 3R = 4R. By T ∝ r^(3/2): T₂/T₁ = (4R/2R)^(3/2) = 2^(3/2) = 2√2. (NCERT Class 11 Physics Chapter 7, page 129.)
Why B is wrong: B is wrong. 4 would be the answer if T ∝ r² (giving (4R/2R)² = 4). Kepler's third law gives T ∝ r^(3/2), so the ratio is 2^(3/2) = 2√2. (Trap: wrong exponent.)
Why C is wrong: C is wrong. 8 = 2³ would result from T ∝ r³, which is not the correct power law. Kepler's third law gives T² ∝ r³.
Why D is wrong: D is wrong. 2 assumes T ∝ r (linear), giving 4R/2R = 2. The correct scaling is T ∝ r^(3/2). (Trap: linear vs 3/2 power.)
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Satellite Time Period: quick recall before you leave
How do you solve a Satellite Time Period question? A 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 129. - 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.
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What to remember before solving Satellite Time Period questions
Which Satellite Time Period formulas do you need for NEET?
1 formula — click to collapse
Kepler's third law
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
More in Gravitation: 4 exam traps and mistakes · 7 formulas · 3 question patterns from its other lessons.
Satellite Time Period questions from past NEET papers
1 question from NEET 2023. Answers verified against NTA official keys. — click to collapse
How does NEET ask about Satellite Time Period?
1 recurring pattern from past papers — click to collapse
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)
Sources
NCERT refs: Class 11 Physics Chapter 7, p.129 | Class 11 Physics Chapter 7, p.139
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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