JEE Mains Physics · Gravitation

Satellites: Orbital Speed, Energy and Mutual Orbits

Gravity supplies the centripetal force, so v = √(GM/r) and T = 2π√(r³/GM); in orbit KE = GMm/2r, PE = −2KE and the total energy is −GMm/2r.

All Gravitation notes JEE Mains Physics strategy

Why this matters

Twenty-eight PYQs, twenty-four of them multiple choice, and two from 2026. Thirteen ask for an orbit's speed, period, height or angular momentum; eight for its energy or the energy to change orbit; seven for bodies that orbit each other. The four numeric-answer questions all sit in the first two groups, and every one of the twenty-eight starts from the same line: gravity equals mv²/r.

Concept 1 of 3: Orbital speed, period and angular momentum

In a circular orbit, gravity is the only force, and it supplies exactly the centripetal force. Setting

GMm/r^{2} = mv^{2}/r

cancels the satellite's mass, so every satellite at the same radius moves at the same speed. Period and angular momentum follow from v and r.

Definition

$v_o = \sqrt{\dfrac{GM}{r}} = \sqrt{\dfrac{gR^{2}}{R + h}}$ , with $r = R + h$ . It is independent of the satellite's mass and falls as $1/\sqrt{r}$ .
Grazing orbit: $v_o = \sqrt{gR} \approx 7.9$ km/s (8 km/s with g = 10).
$T = \dfrac{2\pi r}{v_o} = 2\pi\sqrt{\dfrac{r^{3}}{GM}}$ . Grazing orbit: $T = 2\pi\sqrt{R/g} = \sqrt{\dfrac{3\pi}{G\rho}}$ , about 84 min, set by density alone.
Height from the period: $h = \left(\dfrac{gR^{2}T^{2}}{4\pi^{2}}\right)^{1/3} - R$ .
$L = mv_or = m\sqrt{GMr}$ , so $L \propto m\sqrt{r}$ .
For a central potential $U = -C/r$ , the same balance gives $r \propto 1/v^{2}$ .

Circular orbit of radius r

v_o = \sqrt{\frac{GM}{r}} \qquad T = 2\pi\sqrt{\frac{r^{3}}{GM}} \qquad L = m\sqrt{GMr}

Worked example

A satellite orbits at a height equal to the earth's radius. Find its speed and period. Take

g = 10\ \text{m/s}^{2}

and

R = 6.4 \times 10^{6}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 23 January 2025 · Q97Moderate

Example 1 · Gravitation · Satellites: Orbital Speed, Energy and Mutual Orbits

A satellite of mass

\frac{M}{2}

is revolving around earth in a circular orbit at a height of

\frac{R}{3}

from earth surface. The angular momentum of the satellite is

M\sqrt{\frac{GMR}{x}}

. The value of

x

\_\_\_\_

, where M and

R

are the mass and radius of earth, respectively. ( G is the gravitational constant)

r is R + h

Orbit formulas use the distance from the centre. A satellite 'at a height R' has r = 2R. Putting h in place of r is the commonest error on this page.

Speed does not depend on the satellite's mass; L does

Two satellites in the same orbit have equal speed and period whatever their masses. Their angular momentum and energy scale with mass.

Drill 12 more on orbital speed, period and angular momentum

Concept 2 of 3: Energy of a satellite and the energy to change orbit

In a circular orbit the kinetic energy is exactly half the size of the potential energy. The total is negative, so the satellite is bound. Moving to a higher orbit makes the total less negative, which costs energy, even though the satellite ends up moving more slowly.

Definition

$KE = \dfrac{GMm}{2r}$ , $U = -\dfrac{GMm}{r}$ , $E = -\dfrac{GMm}{2r}$ .
So $U = -2KE$ , $E = -KE$ and $U = 2E$ .
Binding energy (energy to free it from orbit) $= +\dfrac{GMm}{2r}$ .
From radius $r_1$ to $r_2$ : $\Delta E = \dfrac{GMm}{2}\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right)$ .
From the surface (at rest, ignoring the spin) into orbit at r: $\Delta E = GMm\left(\dfrac{1}{R} - \dfrac{1}{2r}\right)$ .
Energies scale with the satellite's mass: $\dfrac{E_A}{E_B} = \dfrac{m_A}{m_B} \cdot \dfrac{r_B}{r_A}$ .

Energies in a circular orbit

KE = \frac{GMm}{2r} \quad U = -\frac{GMm}{r} \quad E = -\frac{GMm}{2r} \qquad \Delta E = \frac{GMm}{2}\left(\frac{1}{r_1} - \frac{1}{r_2}\right)

Worked example

A 200 kg satellite is moved from a circular orbit of radius 2R to one of radius 4R. How much energy is needed? Take

g = 10\ \text{m/s}^{2}

and

R = 6.4 \times 10^{6}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 1 · Q19Moderate

Example 2 · Gravitation · Satellites: Orbital Speed, Energy and Mutual Orbits

Initially a satellite of 100 kg is in a circular orbit of radius

1.5R_{E}

. This satellite can be moved to a circular orbit of radius

3R_{E}

by supplying

\alpha \times10^{6}\text{ }J

of energy. The value of

\alpha

\_\_\_\_

. (Take Radius of Earth

R_{E}= 6 \times10^{6}\text{ }m

and

g = 10m/s^{2}

)

PE is 2E, not E/2

Since E = −GMm/2r and U = −GMm/r, U = 2E. A statement that says PE is half the total energy is false.

Higher orbit: more energy, less speed

Raising a satellite needs energy, yet its kinetic energy falls. The extra energy, and more, goes into potential energy.

Launching is not the same as changing orbits

From the ground at rest, the start energy is −GMm/R with no kinetic energy, so ΔE = GMm(1/R − 1/2r). The orbit-to-orbit formula uses −GMm/2r at both ends.

Drill 7 more on energy of a satellite and the energy to change orbit

Concept 3 of 3: Bodies orbiting each other

When bodies circle under their own attraction, each goes round the common centre of mass. The force uses the distance BETWEEN the bodies; the centripetal term uses each body's own radius about the centre of mass. Keeping these two distances apart is the whole problem.

Definition

Two equal masses m on a circle of radius r: separation $2r$ , so $\dfrac{Gm^{2}}{(2r)^{2}} = \dfrac{mv^{2}}{r}$ , giving $v = \sqrt{\dfrac{Gm}{4r}}$ and $\omega = \sqrt{\dfrac{Gm}{4r^{3}}}$ .
Binary stars $m_1, m_2$ a distance d apart: radii $r_1 = \dfrac{m_2d}{m_1 + m_2}$ , $r_2 = \dfrac{m_1d}{m_1 + m_2}$ , so $m_1r_1 = m_2r_2$ .
Both share one period: $T = 2\pi\sqrt{\dfrac{d^{3}}{G(m_1 + m_2)}}$ .
Four equal masses M on a circle of radius R: net inward pull $\dfrac{GM^{2}}{R^{2}}\left(\dfrac{1}{4} + \dfrac{1}{\sqrt{2}}\right)$ , so $v = \dfrac{1}{2}\sqrt{\dfrac{GM(1 + 2\sqrt{2})}{R}}$ .

Binary system

\frac{Gm_1m_2}{d^{2}} = m_1\omega^{2}r_1 = m_2\omega^{2}r_2 \qquad T = 2\pi\sqrt{\frac{d^{3}}{G(m_1 + m_2)}}

Worked example

Two stars of masses 3m and m, a distance d apart, circle their common centre of mass. Find the period.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 1 · Q7Moderate

Example 3 · Gravitation · Satellites: Orbital Speed, Energy and Mutual Orbits

Two stars of masses

m

and

2m

at a distance d rotate about their common centre of mass in free space. The period of revolution is -

The separation is 2r, not r

Two equal masses on a circle of radius r are 2r apart. Using r in Gm²/r² makes the speed twice too large.

Two different distances in one equation

In a binary, the force uses the separation d; the centripetal term uses the star's own radius about the centre of mass. Each star's radius is the OTHER star's share of the total mass times d.

Drill 6 more on bodies orbiting each other

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Orbital speed, period and angular momentum
Circular orbit of radius r
$v_o = \sqrt{\frac{GM}{r}} \qquad T = 2\pi\sqrt{\frac{r^{3}}{GM}} \qquad L = m\sqrt{GMr}$
Energy of a satellite and the energy to change orbit
Energies in a circular orbit
$KE = \frac{GMm}{2r} \quad U = -\frac{GMm}{r} \quad E = -\frac{GMm}{2r} \qquad \Delta E = \frac{GMm}{2}\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$
Bodies orbiting each other
Binary system
$\frac{Gm_1m_2}{d^{2}} = m_1\omega^{2}r_1 = m_2\omega^{2}r_2 \qquad T = 2\pi\sqrt{\frac{d^{3}}{G(m_1 + m_2)}}$

Watch out for (7)

r is R + h
→ Orbital speed, period and angular momentum
Speed does not depend on the satellite's mass; L does
→ Orbital speed, period and angular momentum
PE is 2E, not E/2
→ Energy of a satellite and the energy to change orbit
Higher orbit: more energy, less speed
→ Energy of a satellite and the energy to change orbit
Launching is not the same as changing orbits
→ Energy of a satellite and the energy to change orbit
The separation is 2r, not r
→ Bodies orbiting each other
Two different distances in one equation
→ Bodies orbiting each other

Test yourself on Gravitation

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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