PhysicsISC Class 11 · Gravitation

Escape Velocity and Satellite Motion: Speed, Period and Energy

Updated 17 October 2026 · Fact-checked

Escape velocity from a planet is v_e = √(2GM/R) = √(2gR); it is √2 times the orbital speed near the surface. A satellite at orbital radius r moves at v = √(GM/r) with period T = 2π√(r³/GM), and its total energy is −GMm/2r.

Understand Escape Velocity and Satellite Motion

Gravitational potential energy. U = −GMm/r, zero at infinity. The work needed to take a mass m from the surface to infinity is GMm/R, so the minimum launch kinetic energy is ½mv_e² = GMm/R, giving v_e = √(2GM/R) = √(2gR). On the Earth, v_e ≈ 11.2 km s⁻¹. It does not depend on the mass of the body or the direction of projection (without air resistance).

Orbital speed. For a circular orbit of radius r = R + h, gravity supplies the centripetal force: GMm/r² = mv²/r, so v = √(GM/r) = √(gR²/(R + h)). Near the surface v_o = √(gR) ≈ 7.9 km s⁻¹, and v_e = √2 v_o. The speed does not depend on the mass of the satellite, and it decreases with the orbit radius.

Period. T = 2πr/v = 2π√(r³/GM), so T² ∝ r³ (Kepler's third law). Near the surface T ≈ 84 minutes. The orbit radius that gives T = 24 h is that of a geostationary satellite, which orbits over the equator in the direction of rotation and appears fixed over a point.

Energy. K = GMm/2r, U = −GMm/r, E = K + U = −GMm/2r = −K. The negative total energy shows the satellite is bound. Energy needed to take it to infinity from the orbit is GMm/2r.

Weightlessness. An astronaut in orbit is in free fall; the apparent weight is zero, though gravity there is still strong.

Escape Velocity and Satellite Motion: formulas and key facts

Escape velocity: $v_{e} = \sqrt (2 GM / R) ∣ v_{e} = \sqrt (2 g R) ∣ v_{e} = \sqrt2 v_{o}$; Earth: about 11.2 km/s.
Orbital speed: $v = \sqrt (GM / r) ∣ v = \sqrt (g R^{2} / (R + h))$; r = R + h.
Period: $T = 2 π \sqrt (r^{3} / GM) ∣ T^{2} \propto r^{3}$; Kepler's third law.
Energies: $K = GM m /2 r ∣ U = - GM m / r ∣ E = - GM m /2 r$; E = −K for a circular orbit.

Escape Velocity and Satellite Motion in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A low-orbit survey satellite

Suppose a satellite circles the Earth at a height of 600 km (take R = 6400 km, g = 9.8 m s⁻²).

Find its orbital speed and time period.

Try it first, then show the solutionSolution

r = 6400 + 600 = 7000 km = 7.0 × 10⁶ m.
v = √(gR²/r) = √(9.8 × 4.096 × 10¹³/7.0 × 10⁶) = √(5.734 × 10⁷) = 7.57 × 10³ m s⁻¹.
T = 2πr/v = 2π × 7.0 × 10⁶/7573 = 5.81 × 10³ s = 96.8 minutes.

Answer: v = 7.57 km s⁻¹; T ≈ 97 minutes.

Memory hook: Higher orbit means lower speed but a longer period.

Tricks to remember Escape Velocity and Satellite Motion

Short memory aids. Say each one aloud once, then use it on a question the same day.

Root two
Escape speed is √2 times the orbital speed at the same radius.
Use it when: Converting between v_e and v_o.
E = −K
In circular orbits, total energy is the negative of kinetic energy and half the potential energy.
Use it when: Energy of a satellite questions.

Escape Velocity and Satellite Motion: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Escape speed from orbital speed · 1 mark · MCQBoard level

The orbital speed of a satellite just above the Earth's surface is about 7.9 km s⁻¹. The escape velocity is therefore nearly:
(a) 5.6 km s⁻¹
(b) 7.9 km s⁻¹
(c) 11.2 km s⁻¹
(d) 15.8 km s⁻¹

Try it first, then show the solutionSolution

v_e = √2 × v_o = 1.414 × 7.9 = 11.2 km s⁻¹.

Answer: (c) 11.2 km s⁻¹

Example 2: Escape velocity of the Earth · 3 marks · Short answerBoard level

Calculate the escape velocity from the Earth's surface using g = 9.8 m s⁻² and R = 6.4 × 10⁶ m.

Try it first, then show the solutionSolution

v_e = √(2gR). (1 mark)
2gR = 2 × 9.8 × 6.4 × 10⁶ = 1.2544 × 10⁸. (1 mark)
v_e = 1.12 × 10⁴ m s⁻¹ = 11.2 km s⁻¹. (1 mark)

Answer: 11.2 km s⁻¹

Example 3: Orbit at a given height · 4 marks · Short answerHarder board level

A weather satellite circles the Earth at a height of 800 km (R = 6400 km, g = 9.8 m s⁻²). Find its orbital speed and period.

Try it first, then show the solutionSolution

r = 7.2 × 10⁶ m; GM = gR² = 9.8 × 4.096 × 10¹³ = 4.014 × 10¹⁴. (1 mark)
v = √(GM/r) = √(5.575 × 10⁷) = 7467 m s⁻¹. (1 mark)
T = 2πr/v = 4.524 × 10⁷/7467. (1 mark)
T = 6059 s = 101 minutes. (1 mark)

Answer: v = 7.47 km s⁻¹; T ≈ 101 minutes.

Example 4: Energy of a satellite · 4 marks · Short answerHarder board level

A 500 kg satellite is in a circular orbit of radius 7.0 × 10⁶ m around the Earth (GM = 4.014 × 10¹⁴ m³ s⁻²). Find its kinetic, potential and total energy and the energy needed to free it from the Earth's attraction.

Try it first, then show the solutionSolution

K = GMm/2r = 4.014 × 10¹⁴ × 500/(1.4 × 10⁷) = 1.434 × 10¹⁰ J. (1 mark)
U = −2K = −2.867 × 10¹⁰ J. (1 mark)
E = K + U = −1.434 × 10¹⁰ J. (1 mark)
Energy needed to reach infinity = +1.434 × 10¹⁰ J. (1 mark)

Answer: K = 1.43 × 10¹⁰ J; U = −2.87 × 10¹⁰ J; E = −1.43 × 10¹⁰ J.

Example 5: Geostationary orbit · 5 marks · Long answerHOTS / competency

Using GM = 4.014 × 10¹⁴ m³ s⁻², R = 6.4 × 10⁶ m, find the radius, height above the surface, and speed of a geostationary satellite (T = 86400 s). State two conditions for the orbit.

Try it first, then show the solutionSolution

r³ = GM T²/4π² = 4.014 × 10¹⁴ × (86400)²/39.48. (1 mark)
86400² = 7.465 × 10⁹, so r³ = 2.996 × 10²⁴/39.48 = 7.59 × 10²² m³. (1 mark)
r = 4.23 × 10⁷ m; h = r − R = 3.59 × 10⁷ m ≈ 36 000 km. (1 mark)
v = 2πr/T = 2π × 4.234 × 10⁷/86400 = 3.08 × 10³ m s⁻¹. (1 mark)
Conditions: the orbit is over the equator and the satellite moves west to east with a period of 24 h. (1 mark)

Answer: r ≈ 4.23 × 10⁷ m; h ≈ 3.6 × 10⁴ km; v ≈ 3.1 km s⁻¹ (with these approximate inputs).

Where students lose marks in Escape Velocity and Satellite Motion

The slips examiners see most often, and what to write instead.

Mistake: Using R instead of R + h for the orbit
Substituting the Earth's radius for the orbit radius.
Do this: Orbit radius r = R + h in v = √(GM/r).
Usually costs: 2 marks.
Mistake: Confusing escape and orbital speed
Quoting 11.2 km/s for a satellite.
Do this: Orbital speed near the surface is √(gR) = 7.9 km/s; escape speed is √2 times that.
Usually costs: 1 to 2 marks.
Mistake: Positive total energy of a satellite
Writing E = +GMm/2r.
Do this: A bound orbit has negative total energy −GMm/2r.
Usually costs: 1 mark.

Exam tips

Write GM = gR² to avoid needing G and M separately.
Convert km to m before substituting.
Remember T² ∝ r³ for ratio questions: you do not need G.
Say that escape velocity is independent of the mass and direction of projection.

Practice questions on Escape Velocity and Satellite Motion

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The orbital speed of a satellite does not depend on:
(a) the mass of the Earth
(b) the radius of the orbit
(c) the mass of the satellite
(d) the value of G
Hint
v = √(GM/r).
Answer
(c) the mass of the satellite
Q2 · 2 marks · Short answerBoard level
A small body has surface gravity 0.6 m s⁻² and radius 2.5 × 10⁵ m. Find the speed needed to escape from it.
Hint
v_e = √(2gR).
Answer
v_e = √(2 × 0.6 × 2.5 × 10⁵) = √(3.0 × 10⁵) = 548 m s⁻¹.
Q3 · 3 marks · Short answerHarder board level
A planet has twice the mass and half the radius of the Earth. Find its escape velocity if the Earth's is 11.2 km s⁻¹.
Hint
v_e ∝ √(M/R).
Answer
v ∝ √(2/0.5) = 2, so v = 22.4 km s⁻¹.
Q4 · 3 marks · Short answerHarder board level
A planet moves round a star in an orbit four times the radius of another planet's orbit around the same star. If the inner planet's year is 1 earth year, find the outer planet's year.
Hint
T² ∝ r³.
Answer
T² = 4³ = 64, so T = 8 years.
Q5 · 4 marks · Short answerHarder board level
Find the minimum energy needed to put a 1000 kg satellite into a circular orbit 600 km above the Earth, starting from rest at the surface (GM = 4.014 × 10¹⁴ m³ s⁻², R = 6.4 × 10⁶ m). Ignore the rotation of the Earth.
Hint
ΔE = E_orbit − E_surface = GMm(1/R − 1/2r).
Answer
ΔE = 4.014 × 10¹⁷ × (1/6.4 × 10⁶ − 1/1.4 × 10⁷) = 4.014 × 10¹⁷ × 8.482 × 10⁻⁸ = 3.4 × 10¹⁰ J.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose astronauts live in a station orbiting at 400 km above the Earth (R = 6400 km, GM = 4.014 × 10¹⁴ m³ s⁻²).
(i) Find the orbital speed. (1 mark)
(ii) Find the time period. (1 mark)
(iii) How many orbits do they complete in a day? (1 mark)
(iv) Explain why they feel weightless although g there is about 8.7 m s⁻². (1 mark)
Hint
r = 6.8 × 10⁶ m.
Answer
(i) v = √(4.014 × 10¹⁴/6.8 × 10⁶) = 7.68 km s⁻¹. (ii) T = 2π × 6.8 × 10⁶/7683 = 5561 s = 92.7 min. (iii) 86400/5561 = 15.5 orbits. (iv) Both the station and the astronauts are in free fall with the same acceleration, so no contact force acts between them; g there = 9.8 × (6400/6800)² = 8.68 m s⁻².

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Escape velocity formula?Show answer
v_e = √(2gR) = √(2GM/R).
2.Orbital speed at radius r?Show answer
v = √(GM/r).
3.Total energy of a satellite?Show answer
−GMm/2r.
4.Period of a geostationary satellite?Show answer
24 hours.

Escape Velocity and Satellite Motion: frequently asked questions

Is escape velocity the speed needed to orbit?

No. Escape velocity takes a body to infinity. A body needs only √(gR) to orbit just above the surface, which is 1/√2 times the escape speed.

Why do satellites in higher orbits move slower?

Gravity is weaker farther away, so less centripetal acceleration is needed, which means a lower speed: v ∝ 1/√r.

Why must a geostationary satellite be over the equator?

Its orbit must have its centre at the Earth's centre and rotate with the Earth; only an equatorial orbit stays above the same point on the surface.