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. 137Motion of Satellite
8 MCQs4 revision cards9-step worked example
Source: NCERT GravitationOfficial key: NTA-verifiedLast updated: 25 Sep 2026
Motion of Satellite, explained for NEET
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 138). 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.
Can you answer these Motion of Satellite 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)
Show answer and why every option is right or wrong
Answer: B. B is correct. For a near-surface orbit, v₀ = √(gR) = √(9.8 × 6.4 × 10⁶) ≈ 7.9 × 10³ m/s = 7.9 km/s. This is a standard result stated in NCERT Class 11 Physics Chapter 7, page 138.
Why A is wrong: A is wrong because 6.4 km/s confuses the Earth's radius value (6400 km) with the orbital speed.
Why C is wrong: C is wrong because 9.8 km/s transplants the numerical value of g (m/s²) directly as a speed — the formula requires √(gR), not g itself.
Why D is wrong: D is wrong because 11.2 km/s is the escape velocity from Earth's surface (v_e = √(2gR)), not the orbital velocity. The orbital velocity is smaller by a factor of √2.
MCQ 2Easy RecallPractice
The time period of a satellite in a circular orbit close to Earth's surface is approximately:
Show answer and why every option is right or wrong
Answer: C. C is correct. For an orbit just above the surface, T = 2π√(R/g) = 2π√(6.37 × 10⁶ / 9.8) ≈ 5.07 × 10³ s ≈ 84.4 minutes. This is the minimum orbital period for an Earth satellite (NCERT Class 11 Physics Chapter 7, page 138).
Why A is wrong: A is wrong because 24 hours is the period of a geostationary satellite at ~36,000 km altitude, not a near-surface orbit.
Why B is wrong: B is wrong because 12 hours corresponds roughly to a GPS satellite orbit (~20,200 km altitude), far above the surface.
Why D is wrong: D is wrong because 60 minutes would require an orbit radius smaller than Earth's radius — physically impossible for a surface-skimming orbit.
MCQ 3Easy RecallPractice
The ratio of escape velocity to orbital velocity at the same point near Earth's surface is:
Show answer and why every option is right or wrong
Answer: B. B is correct. v_e = √(2gR) and v₀ = √(gR), so v_e/v₀ = √2. This ratio holds at any radius — it is a direct consequence of the energy conditions for bound versus unbound orbits.
Why A is wrong: A is wrong because equating escape and orbital velocities would mean a circular orbit already has enough energy to escape, which contradicts the negative total energy of bound orbits.
Why C is wrong: C is wrong because v_e/v₀ = 2 would require v_e = 2√(gR), which gives v_e² = 4gR instead of 2gR — double the actual kinetic energy needed for escape.
Why D is wrong: D is wrong because this inverts the ratio. v_orbital is smaller than v_escape, so v_e/v₀ > 1, not < 1.
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:
Show answer and why every option is right or wrong
Answer: D. D is correct. Total energy E = −GMm/(2r). At radius 2r, E' = −GMm/(2 × 2r) = −GMm/(4r) = E/2. Since E is negative, E/2 is less negative — the satellite has more energy (less tightly bound) in the higher orbit, which is physically consistent.
Why A is wrong: A is wrong because 2E would mean doubling the (negative) total energy, making the satellite more tightly bound at a larger radius — this contradicts the inverse dependence of |E| on orbit radius.
Why B is wrong: B is wrong because 4E would make the binding energy four times stronger at double the radius. Energy goes as 1/r, so moving farther reduces binding — it does not increase it.
Why C is wrong: C is wrong because E/4 would result from an inverse-square dependence of E on r. Total energy depends as 1/r (not 1/r²), so doubling r halves |E|, giving E/2.
MCQ 5Direct ApplicationPractice
For a satellite in a stable circular orbit, which relation is correct?
Show answer and why every option is right or wrong
Answer: A. A is correct. KE = GMm/[2(R+h)] and PE = −GMm/(R+h), so KE = −PE/2. Equivalently, E = KE + PE = −KE, confirming the virial theorem for a 1/r² force. (NCERT Class 11 Physics Chapter 7, page 140.)
Why B is wrong: B is wrong because KE = PE would give total energy E = 2KE > 0, meaning an unbound orbit — contradicting the premise of a stable circular orbit.
Why C is wrong: C is wrong because KE = −2PE would give KE = 2GMm/(R+h) and total E = KE + PE = GMm/(R+h) > 0 — again an unbound state, not a circular orbit.
Why D is wrong: D is wrong because KE = PE/2 gives KE = −GMm/[2(R+h)] < 0. Kinetic energy cannot be negative. This error arises from dropping the sign of PE.
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:
Show answer and why every option is right or wrong
Answer: C. C is correct. ΔE = E_final − E_initial = −GMm/(2 × 3R) − [−GMm/(2 × 2R)] = −GMm/(6R) + GMm/(4R) = GMm[1/4 − 1/6]/(R) = GMm × (1/12)/R = GMm/(12R). The energy input must equal this positive value.
Why A is wrong: A is wrong because GMm/(3R) is the magnitude of the potential energy at 3R alone, not a difference between the energies of the two orbits.
Why B is wrong: B is wrong because GMm/(6R) equals the magnitude of the total energy at 3R alone. The question asks for the difference between two orbital energies, not the energy at the final orbit.
Why D is wrong: D is wrong because GMm/(2R) ignores that the satellite already has orbital energy at 2R. The required energy is the difference in total energies, which is much smaller than GMm/(2R).
MCQ 7Concept TrapPractice
An astronaut inside an orbiting space station feels weightless because:
Show answer and why every option is right or wrong
Answer: D. D is correct. Both the astronaut and the station accelerate toward Earth at the same rate (free fall), so there is no normal contact force between them. At ISS altitude (~400 km), g is still about 8.7 m/s² — gravity has not vanished.
Why A is wrong: A is wrong because g at 400 km altitude is approximately 8.7 m/s², roughly 89% of surface g. Gravitational acceleration is far from zero.
Why B is wrong: B is wrong because 'centripetal force' is not a separate force cancelling gravity — gravity IS the centripetal force. There is no additional outward force to balance. The apparent weightlessness comes from shared free fall, not force cancellation.
Why C is wrong: C is wrong because Earth's gravitational field extends to infinity (decreasing as 1/r²). At ISS altitude, the field strength is substantial. There is no boundary beyond which the field vanishes.
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:
Show answer and why every option is right or wrong
Answer: A. A is correct. Energy on the surface: E_i = KE + PE = 0 + (−GMm/R) = −GMm/R. Energy in orbit at r = 2R: E_f = −GMm/(2 × 2R) = −GMm/(4R). Energy supplied = E_f − E_i = −GMm/(4R) − (−GMm/R) = −GMm/(4R) + GMm/R = 3GMm/(4R). This includes both the PE raise and the orbital KE acquisition.
Why B is wrong: B is wrong because GMm/(4R) equals the magnitude of the orbital energy at 2R, not the energy difference from the surface. This error typically comes from forgetting to subtract the surface energy (−GMm/R).
Why C is wrong: C is wrong because GMm/(2R) accounts only for the change in gravitational PE from R to 2R (ΔPE = −GMm/(2R) + GMm/R = GMm/(2R)) while neglecting the orbital kinetic energy the satellite must acquire.
Why D is wrong: D is wrong because 5GMm/(4R) would result from adding |PE at surface| and |E at orbit| instead of taking their difference. The satellite doesn't need energy equal to |E_surface| + |E_orbit|; it needs only the gap between the two energy states.
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Motion of Satellite: quick recall before you leave
How do you solve a Motion of Satellite question? A 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).)
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What to remember before solving Motion of Satellite questions
Which Motion of Satellite formulas do you need for NEET?
1 formula — click to collapse
Orbital velocity for circular orbit
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
More in Gravitation: 4 exam traps and mistakes · 7 formulas · 4 question patterns from its other lessons.
Motion of Satellite questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
NCERT refs: Class 11 Physics Chapter 7, p.138
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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