Formula
Energy of a satellite
Total mechanical energy E = -G M m / (2(R+h)) = -KE = U/2 (virial relations for circular orbit). Negative E means the satellite is bound; making E ≥ 0 escapes.
-- NCERT Class 11 Physics, Ch. 7, p. 140Satellite Energy
8 MCQs10 revision cards9-step worked example
Source: NCERT GravitationPYQ coverage: NEET 2024Official key: NTA-verifiedLast updated: 25 Sep 2026
Satellite Energy, explained for NEET
The energy of an orbiting satellite is a bound-state accounting problem — and the trap that costs marks is forgetting one side of the ledger.
The core relationship. For a satellite of mass m in a circular orbit at radius r = R + h around Earth (mass M):
- Gravitational PE: U = −GMm/r
- Orbital KE: K = GMm/(2r)
- Total mechanical energy: E = K + U = −GMm/(2r)
The total energy is exactly −K, or equivalently U/2. This is the virial theorem result for inverse-square forces. A bound satellite always has E < 0. If energy is added until E = 0, the satellite escapes — it reaches the threshold of an unbound orbit.
The high-frequency trap. When a question asks "how much energy is needed to move a satellite from orbit 1 to orbit 2," aspirants commonly compute only the PE change (ΔU) and forget that the satellite must also change its orbital speed. The correct answer is the change in total energy: ΔE = E₂ − E₁. Since E includes both KE and PE, you cannot ignore either. Similarly, the energy to launch a satellite from rest on the surface to a circular orbit at height h is not just the PE gain — you must also supply the orbital KE.
Sign discipline. E is negative and its magnitude decreases as r increases (the satellite becomes less tightly bound at higher orbits). "More energy" means E becomes less negative — closer to zero, not more negative.
NCERT Class 11 Physics, Chapter 7 (Gravitation), page 140 derives the satellite energy formula by combining the orbital velocity condition with the PE expression.
Can you answer these Satellite Energy 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
For a satellite in a stable circular orbit around Earth, which statement about its total mechanical energy E is correct?
Show answer and why every option is right or wrong
Answer: C. A bound satellite always has negative total energy. E = −GMm/(2r), which is negative for all finite r. NCERT Class 11 Physics, Chapter 7, page 140 states this directly.
Why A is wrong: A positive total energy would mean the satellite is unbound (hyperbolic trajectory), not in a stable orbit.
Why B is wrong: Zero total energy corresponds to the escape condition — the satellite would fly off to infinity, not remain in orbit.
Why D is wrong: For any circular orbit at finite altitude, E = −GMm/(2r) is strictly negative. There is no altitude that makes it positive while maintaining a bound orbit.
MCQ 2Easy RecallPractice
For a satellite in circular orbit, what is the relationship between its kinetic energy K and total mechanical energy E?
Show answer and why every option is right or wrong
Answer: B. In a circular orbit, K = GMm/(2r) and E = −GMm/(2r), so E = −K. This is a direct consequence of the virial theorem for inverse-square forces. NCERT Class 11 Physics, Chapter 7, page 140.
Why A is wrong: E = K would imply zero potential energy, which contradicts U = −GMm/r ≠ 0 for any finite orbit.
Why C is wrong: E = 2K would give E = GMm/r (positive), but total energy of a bound orbit is always negative.
Why D is wrong: E = −2K would give E = −GMm/r = U, ignoring the kinetic energy entirely.
MCQ 3Easy RecallPractice
For a satellite of mass m in a circular orbit of radius r around a planet of mass M, the gravitational potential energy is U = −GMm/r. What is the ratio |U|/K, where K is the orbital kinetic energy?
Show answer and why every option is right or wrong
Answer: C. K = GMm/(2r), so |U| = GMm/r = 2K, giving |U|/K = 2. NCERT Class 11 Physics, Chapter 7, page 140.
Why A is wrong: |U|/K = 1/2 reverses the ratio — that would be K/|U|, not |U|/K.
Why B is wrong: |U|/K = 1 would mean K = |U| = GMm/r, but orbital KE is GMm/(2r), half that value.
Why D is wrong: |U|/K = 4 has no basis in the circular orbit energy expressions.
MCQ 4Direct ApplicationPractice
A satellite orbits Earth in a circular orbit of radius r. If it is moved to a new circular orbit of radius 2*r*, what is the ratio of its new total energy E₂ to the original total energy E₁?
Show answer and why every option is right or wrong
Answer: B. E = −GMm/(2r), so E₂/E₁ = [−GMm/(2·2r)] / [−GMm/(2r)] = 1/2. The magnitude of total energy halves when the orbital radius doubles. NCERT Class 11 Physics, Chapter 7, page 140.
Why A is wrong: 1/4 would apply if E ∝ 1/r², but satellite total energy is proportional to 1/r, not 1/r².
Why C is wrong: A ratio of 2 would mean E₂ is more negative (more tightly bound) at a larger orbit, which contradicts the fact that higher orbits are less tightly bound.
Why D is wrong: A ratio of 4 has no basis — neither E ∝ 1/r nor E ∝ 1/r² gives a factor of 4 for doubling r.
MCQ 5Direct ApplicationPractice
A satellite of mass m is in a circular orbit at height h = R above Earth's surface (orbit radius = 2*R*). What is its total mechanical energy? (Take g as surface gravitational acceleration, R as Earth's radius.)
Show answer and why every option is right or wrong
Answer: B. E = −GMm/(2·2R) = −GMm/(4R). Since GM = gR², this becomes −gR²m/(4R) = −mgR/4. NCERT Class 11 Physics, Chapter 7, page 140.
Why A is wrong: −mgR/2 corresponds to the total energy at the surface (orbit radius R), not at height h = R (orbit radius 2R).
Why C is wrong: −mgR/8 would correspond to orbit radius 4R (height 3R), not 2R.
Why D is wrong: −mgR ignores the factor of 2 in the denominator of the satellite energy formula and the orbit radius of 2R.
MCQ 6Direct ApplicationPractice
A satellite is in a circular orbit with total energy E. What additional energy must be supplied to make it escape to infinity?
Show answer and why every option is right or wrong
Answer: B. At infinity with zero velocity, total energy = 0. The satellite's current energy is E (which is negative). Energy to supply = 0 − E = −E (a positive quantity since E < 0). NCERT Class 11 Physics, Chapter 7, page 140.
Why A is wrong: Supplying energy equal to E (a negative number) would mean removing energy, making the satellite more tightly bound, not freeing it.
Why C is wrong: 2E is negative (since E < 0), so supplying 2E would decrease the total energy further — the opposite of escape.
Why D is wrong: −2E would overshoot: the satellite would escape with residual kinetic energy equal to |E|, not just barely escape.
MCQ 7CalculationPractice
A satellite of mass m rests on Earth's surface. What minimum energy must be supplied to place it in a circular orbit at height h = R above the surface? (Neglect air resistance. Use g for surface gravity, R for Earth's radius.)
Show answer and why every option is right or wrong
Answer: A. On the surface (at rest), E_initial = PE only = −GMm/R = −mgR. In orbit at radius 2R, E_final = −GMm/(2·2R) = −mgR/4. Energy required = E_final − E_initial = −mgR/4 − (−mgR) = (3/4)mgR. This accounts for both the PE change and the orbital KE. NCERT Class 11 Physics, Chapter 7, page 140.
Why B is wrong: (5/4)mgR results from taking the PE rise as mgh = mgR (uniform g, wrong for h = R) and then adding the orbital KE, mgR/4. The exact PE rise is only mgR/2.
Why C is wrong: (1/2)mgR is the exact potential-energy rise, −GMm/(2R) − (−GMm/R) = mgR/2, but it leaves out the orbital kinetic energy GMm/(4R) = mgR/4 the satellite must also be given.
Why D is wrong: (1/4)mgR is just the magnitude of the final orbital energy — it ignores the initial surface energy entirely.
MCQ 8CalculationPractice
Two satellites A and B of masses m and 2*m* orbit Earth in circular orbits of radii 2*R* and 4*R* respectively. What is the ratio of total energy of A to that of B, i.e., E_A/E_B?
Show answer and why every option is right or wrong
Answer: A. E_A = −GMm/(2·2R) = −GMm/(4R). E_B = −GM(2m)/(2·4R) = −GMm/(4R). Therefore E_A/E_B = 1. Doubling both mass and orbit radius leaves the total energy unchanged. NCERT Class 11 Physics, Chapter 7, page 140.
Why B is wrong: A ratio of 2 results from ignoring B's doubled mass — computing E_B as if mass were m, not 2m.
Why C is wrong: A ratio of 1/2 comes from accounting for B's doubled mass but not its doubled orbit radius, giving E_B = 2 × E_A.
Why D is wrong: A ratio of 4 has no basis in the formula E = −GMm/(2r); no combination of ×2 mass and ×2 radius produces a factor of 4.
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Satellite Energy: quick recall before you leave
How do you solve a Satellite Energy question? A worked example
Pattern: Energy required to launch a satellite to orbit (NEET pattern: satellite orbit energy — observed NEET 2024)
- 1
Given
• Mass of satellite: m = 500 kg• Altitude: h = R = 6.4 × 10⁶ m• Orbit radius: r = R + h = 2R = 1.28 × 10⁷ m• Surface gravity: g = 9.8 m/s² (exact, problem-defined)• Earth radius: R = 6.4 × 10⁶ m
- 2
Required
Minimum energy E_input to move satellite from rest on surface to circular orbit at radius 2R.
- 3
Concept
The energy input equals the change in total mechanical energy: E_input = E_final − E_initial. On the surface at rest, E_initial = U_surface = −GMm/R. In circular orbit, E_final = −GMm/(2·2R) = −GMm/(4R). Both KE (orbital) and PE changes are captured by this total-energy approach.
- 4
Formula
E_input = E_final − E_initial = −GMm/(4R) − (−GMm/R) = (3/4) × GMm/R
Using GM = gR²:
E_input = (3/4) × gR²m/R = (3/4)mgR - 5
Substitution
E_input = (3/4) × 500 × 9.8 × 6.4 × 10⁶
- 6
Calculation
• 500 × 9.8 = 4900• 4900 × 6.4 × 10⁶ = 3.136 × 10¹⁰• (3/4) × 3.136 × 10¹⁰ = 2.352 × 10¹⁰ J
Note: the constants g = 9.8 m/s² and R = 6.4 × 10⁶ m are treated as exact (problem-defined values), so they do not limit significant figures. The factor 3/4 is an exact rational number. The answer precision is set by the satellite mass (3 significant figures). - 7
Final answer
E_input = 2.35 × 10¹⁰ J ≈ 23.5 GJ (to 3 significant figures).
- 8
Common trap
Computing only the PE change: ΔU = −GMm/(2R) − (−GMm/R) = GMm/(2R), which gives (1/2)mgR ≈ 1.57 × 10¹⁰ J. This is wrong because it omits the orbital kinetic energy the satellite must acquire. The correct method uses total energy, which automatically includes both KE and PE.
- 9
Similar NEET-style question
"A satellite of mass 200 kg is to be placed in a circular orbit at height 2R above Earth's surface. Find the minimum energy required." (Answer: E = (5/6)mgR — work this out using E_final = −GMm/(2·3R) and E_initial = −GMm/R.)
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What to remember before solving Satellite Energy questions
Which Satellite Energy formulas do you need for NEET?
1 formula — click to collapse
Satellite total mechanical energy
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)
More in Gravitation: 4 exam traps and mistakes · 7 formulas · 3 question patterns from its other lessons.
Satellite Energy questions from past NEET papers
1 question from NEET 2024. Answers verified against NTA official keys. — click to collapse
NEET 2024Revised key
16R 2GmM
23R GmM
32R GmM
43R
NTA Answer: Option 1(revised_final)
How does NEET ask about Satellite Energy?
1 recurring pattern from past papers — click to collapse
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
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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