PhysicsCBSE Class 11 · Gravitation

Satellites, Orbital Energy and Escape Speed Class 11

Q: Why is a satellite's total energy negative?

Because it is gravitationally bound; it would need positive energy to escape.

Q: Does a satellite need fuel to stay in orbit?

No; in a vacuum gravity alone provides the centripetal force, and fuel is only needed to change orbits.

Q: Why does escape speed not depend on mass?

The mass of the projectile cancels from the energy equation.

Q: Can a geostationary satellite be placed over the poles?

No; it must orbit in the equatorial plane to stay fixed above one point.

Q: Why do astronauts feel weightless?

They are in free fall with the spacecraft, so there is no normal reaction.

Updated 17 October 2026 · Fact-checked

A satellite in a circular orbit of radius r has speed v = √(GM/r) and period T = 2π√(r³/GM). Its kinetic energy is +GMm/2r, potential energy −GMm/r and total energy −GMm/2r. The escape speed from a surface is √(2GM/R) = √(2gR), and from a circular orbit the extra speed needed is a factor √2 − 1 of the orbital speed.

Notice that U is twice as deep as K is tall, so total E = −K is negative: the satellite is bound until it gains energy to reach zero.

Illustration drawn for QuestionBuddy.

Understand Satellites, Orbital Energy and Escape Speed

Orbital motion. Gravity supplies the centripetal force: $\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r}$ , so
$ $v = \frac{GM}{r}, T = \frac{2 π r}{v} = 2 π \frac{r ^{3}}{GM}, T^{2} \propto r^{3} .$ $
Using $GM = g R^{2}$ , $v = R g / r$ . Near the surface ( $r \approx R$ ): $v = g R \approx 7.9 km/s$ and $T \approx 84$ min. The speed is lower in a higher orbit. A satellite is weightless (in free fall) but not free of gravity.

Geostationary satellite. Period 24 h, orbit in the equatorial plane, moving west to east, at one fixed height (about $3.6 \times 1 0^{7}$ m above the surface, found from $r^{3} = GM T^{2} /4 π^{2}$ ). It appears fixed over a point on the equator.

Energy. For a circular orbit:
$ $K = \frac{GM m}{2 r}, U = - \frac{GM m}{r}, E = K + U = - \frac{GM m}{2 r} = - K = \frac{1}{2} U .$ $
The total energy is negative: the satellite is bound. The binding energy is $+ GM m /2 r$ . If a satellite loses energy by drag it moves to a lower orbit and its speed *increases*, since $K = - E$ .

Escape speed. To reach infinity with zero speed, total energy must be at least zero: $\frac{1}{2} m v_{e}^{2} - \frac{GM m}{R} = 0$ , so
$ $v_{e} = \frac{2 GM}{R} = 2 g R = 2 v_{or bi t (s u r f a ce)} .$ $
For the Earth $v_{e} \approx 11.2$ km/s; it is independent of the mass and direction of projection. From a circular orbit of radius $r$ , the escape speed is $2 v_{or b}$ , so the extra kinetic energy required equals $GM m /2 r$ (the binding energy).

Launching. The energy needed to put a mass $m$ into an orbit of radius $r$ from rest on the surface is $E_{r} - E_{R} = \frac{GM m}{R} - \frac{GM m}{2 r}$ (ignoring the Earth's rotation).

Satellites, Orbital Energy and Escape Speed: formulas and key facts

Orbital speed: $v = \frac{GM}{r} = R \frac{g}{r}$; Circular orbit.
Period: $T = 2 π \frac{r ^{3}}{GM}$; $T^{2} \propto r^{3}$ .
Total energy: $E = - \frac{GM m}{2 r}$; Bound orbit.
Kinetic and potential: $K = \frac{GM m}{2 r}, U = - \frac{GM m}{r}$; $E = - K$ .
Escape speed: $v_{e} = \frac{2 GM}{R} = 2 g R$; From the surface.
Energy to launch into orbit: $Δ E = \frac{GM m}{R} - \frac{GM m}{2 r}$; From rest on the surface.

How to solve Satellites, Orbital Energy and Escape Speed questions

Equate gravity to centripetal force for an orbit; use energy conservation (zero total energy) for escape.

1Write GM = gR² if only g and R are given.
2For circular orbits, find v from GM/r and T from 2πr/v.
3For energy, write K, U and E = −GMm/2r with the orbit radius measured from the centre.
4For escape, set total energy to zero at the surface or in the orbit.
5For a launch, subtract the initial energy (−GMm/R on the ground) from the final energy.
6Check signs: bound energies are negative.

Satellites, Orbital Energy and Escape Speed 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-Earth satellite

Suppose a satellite orbits at 7.5 km/s.

Is that consistent with an orbit close to the surface (g = 9.8 m/s², R = 6400 km)?

Try it first, then show the solutionSolution

Near-surface orbital speed = √(9.8 × 6.4 × 10⁶) = 7.9 km/s.
7.5 km/s is smaller, so the orbit is somewhat higher than the surface.

Answer: Yes, a slightly higher orbit.

Memory hook: Higher orbits mean lower speeds.

A communications dish

Suppose a dish must point at a satellite that appears fixed in the sky above the equator.

What are the satellite's period and sense of rotation?

Try it first, then show the solutionSolution

It must keep pace with the Earth: period 24 h.
It moves west to east in the equatorial plane.

Answer: 24 h, west to east, over the equator.

Memory hook: Geostationary means fixed relative to the ground.

Tricks to remember Satellites, Orbital Energy and Escape Speed

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

Energy ladder
K = +GMm/2r, E = −GMm/2r, U = −GMm/r: E = −K = U/2.
Use it when: Orbital energy questions.
Escape = √2 × orbital
v_escape = √2 v_orbit at the same r.
Use it when: Boost questions.
T² ∝ r³
Kepler's law applies to every circular orbit.
Use it when: Comparing satellites.
Drag speeds you up
Lower orbit means larger v.
Use it when: Conceptual items.

Satellites, Orbital Energy and Escape Speed: solved examples

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

Example 1: Near-surface satellite · 2 marks · Very short answerBoard level

Find the orbital speed and period of a satellite that skims the Earth's surface (g = 9.8 m/s², R = 6.4 × 10⁶ m).

Try it first, then show the solutionSolution

v = √(gR) = √(9.8 × 6.4 × 10⁶) ≈ 7.9 × 10³ m/s. (1)
T = 2πR/v = 2π × 6.4 × 10⁶/7920 ≈ 5.1 × 10³ s ≈ 85 min. (1)

Answer: About 7.9 km/s and 85 min.

Example 2: Escape from the Moon and the Earth · 3 marks · Short answerHarder board level

The Moon has g = 1.6 m/s² and radius 1.74 × 10⁶ m. Find its escape speed and compare it with the Earth's (11.2 km/s). What does this say about the Moon's atmosphere?

Try it first, then show the solutionSolution

v_e = √(2gR) = √(2 × 1.6 × 1.74 × 10⁶) ≈ 2.36 × 10³ m/s. (1)
Ratio = 11.2/2.36 ≈ 4.7: the Earth's is almost five times larger. (1)
With a low escape speed, gas molecules can reach it and escape, so the Moon cannot retain an atmosphere. (1)

Answer: 2.4 km/s; about one-fifth of the Earth's.

Example 3: Energy of an orbiting satellite · 3 marks · Short answerHarder board level

A 500 kg satellite orbits at r = 2R (g = 9.8 m/s², R = 6.4 × 10⁶ m). Find its kinetic energy, potential energy, total energy and binding energy.

Try it first, then show the solutionSolution

GMm/r = gR²m/2R = gRm/2 = 9.8 × 6.4 × 10⁶ × 500/2 = 1.568 × 10¹⁰ J. (1)
U = −1.57 × 10¹⁰ J; K = +GMm/2r = 7.84 × 10⁹ J. (1)
E = −7.84 × 10⁹ J; binding energy = +7.84 × 10⁹ J. (1)

Answer: K = 7.84 × 10⁹ J; U = −1.57 × 10¹⁰ J; E = −7.84 × 10⁹ J.

Example 4: Geostationary orbit · 4 marks · Long answerHarder board level

Using g = 9.8 m/s², R = 6.4 × 10⁶ m and a period of 24 h, find the radius of a geostationary orbit, its height above the surface and the satellite's speed.

Try it first, then show the solutionSolution

GM = gR² = 4.01 × 10¹⁴ m³/s². (1)
r³ = GMT²/4π² = 4.01 × 10¹⁴ × (86400)²/39.5 = 7.59 × 10²² m³, so r ≈ 4.23 × 10⁷ m. (1)
Height = 4.23 × 10⁷ − 6.4 × 10⁶ ≈ 3.6 × 10⁷ m. (1)
v = 2πr/T = 2π × 4.23 × 10⁷/86400 ≈ 3.08 × 10³ m/s. (1)

Answer: r ≈ 4.2 × 10⁷ m; height ≈ 3.6 × 10⁷ m; v ≈ 3.1 km/s.

Example 5: Boosting a satellite · 4 marks · Long answerHOTS / competency

A 500 kg satellite is in a circular orbit of radius 2R (g = 9.8 m/s², R = 6.4 × 10⁶ m). A thruster suddenly increases its speed by 20% along its path. (a) Will it escape? (b) Find its new total energy and compare with the old. (c) How much extra energy would be needed to just escape from the original circular orbit?

Try it first, then show the solutionSolution

(a) New speed = 1.2 v_orb < √2 v_orb = 1.41 v_orb, so it does not escape. (1)
(b) New E = ½m(1.44v²) − GMm/r = (0.72 − 1) GMm/r = −0.28 GMm/r = −0.28 × 1.568 × 10¹⁰ ≈ −4.4 × 10⁹ J. (1)
Old E = −7.84 × 10⁹ J, so the new orbit is an ellipse with 56% of the old binding energy. (1)
(c) Extra energy to escape from the original orbit = GMm/2r = 7.84 × 10⁹ J. (1)

Answer: (a) no; (b) −4.4 × 10⁹ J (old −7.84 × 10⁹ J); (c) 7.84 × 10⁹ J.

Where students lose marks in Satellites, Orbital Energy and Escape Speed

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

Mistake: Using the height instead of the orbital radius in GM/r.
Measuring from the surface.
Do this: r is measured from the centre: r = R + h.
Usually costs: The answer mark.
Mistake: Writing the total energy as +GMm/2r.
Dropping the sign.
Do this: E = −GMm/2r: bound orbits have negative energy.
Usually costs: 1 to 2 marks.
Mistake: Saying the satellite has no weight because there is no gravity.
Confusing free fall with zero gravity.
Do this: It is in free fall; gravity provides the centripetal force.
Usually costs: Concept mark.
Mistake: Claiming escape speed depends on the mass of the projectile.
Thinking heavy bodies need more speed.
Do this: v_e = √(2GM/R) does not contain m.
Usually costs: 1 mark.

Exam tips

Start with "gravity provides the centripetal force" for orbit questions.
Quote GM = gR² when g and R are given.
State the sign of every energy.
For launch energy, include the initial energy −GMm/R.
Geostationary: mention the plane, direction and period.

Practice questions on Satellites, Orbital Energy and Escape Speed

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
To escape from a circular orbit, a satellite's speed must be increased by a factor of
(a) 2 (b) √2 (c) 1/√2 (d) 4
Hint
v_esc = √2 v_orb.
Answer
(b) √2: the speed must reach √2 times the orbital speed.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): When a satellite loses energy due to atmospheric drag, its speed increases.
Reason (R): The total energy −GMm/2r becomes more negative in a lower orbit, where the kinetic energy GMm/2r is larger.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
K = −E.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
One satellite orbits at 9 times the orbital radius of another around the same planet. Find the ratios of their periods and orbital speeds.
Hint
T² ∝ r³, v ∝ 1/√r.
Answer
T ratio = 9^{3/2} = 27 : 1; speed ratio = 1 : 3.
Q4 · 3 marks · Short answerHarder board level
A planet has one-tenth the Earth's mass and half its radius. Find its escape speed (Earth: 11.2 km/s).
Hint
v_e ∝ √(M/R).
Answer
v_e = 11.2 × √(0.1/0.5) = 11.2 × 0.447 ≈ 5.0 km/s.
Q5 · 4 marks · Long answerHOTS / competency
A moon orbits a planet at a radius of 6.0 × 10⁸ m with a period of 4 days. (a) Find the mass of the planet (G = 6.67 × 10⁻¹¹ N m²/kg²). (b) Find g on its surface and the escape speed if its radius is 7.0 × 10⁷ m.
Hint
M = 4π²r³/GT².
Answer
T = 3.456 × 10⁵ s. (a) M = 4π² × 2.16 × 10²⁶/(6.67 × 10⁻¹¹ × 1.194 × 10¹¹) ≈ 1.07 × 10²⁷ kg. (b) g = GM/R² ≈ 14.6 m/s²; v_e = √(2GM/R) ≈ 45 km/s.

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.Orbital speed?Show answer
√(GM/r).
2.Total energy of a satellite?Show answer
−GMm/2r.
3.Escape speed of the Earth?Show answer
About 11.2 km/s.
4.Relation of K and E?Show answer
E = −K.
5.Geostationary period?Show answer
24 hours.
6.Kepler's third law?Show answer
T² ∝ r³.

Satellites, Orbital Energy and Escape Speed: frequently asked questions

Why is a satellite's total energy negative?

Because it is gravitationally bound; it would need positive energy to escape.

Does a satellite need fuel to stay in orbit?

No; in a vacuum gravity alone provides the centripetal force, and fuel is only needed to change orbits.

Why does escape speed not depend on mass?

The mass of the projectile cancels from the energy equation.

Can a geostationary satellite be placed over the poles?

No; it must orbit in the equatorial plane to stay fixed above one point.

Why do astronauts feel weightless?

They are in free fall with the spacecraft, so there is no normal reaction.