Satellite, Planetary Motion And Kepler’s Law

PhysicsGravitationFor JEE aspirants

A satellite in a stable circular orbit around a planet of mass $M$ at radius $r$ has orbital speed $v_{0} = GM / r$ and time period $T = 2 π r^{3} / GM$ , because gravity supplies exactly the centripetal force required. Kepler's three laws generalise this to elliptical planetary orbits: planets move in ellipses with the Sun at one focus, sweep equal areas in equal times, and satisfy $T^{2} \propto a^{3}$ . The escape velocity from Earth's surface is $v_{e} = 2 g R \approx 11.2 km/s$ . This concept covers orbital motion, energy in orbits, geostationary and polar satellites, weightlessness, and Kepler's laws for JEE Mains and NEET.

Key Formulas - Quick Reference

Orbital speed at radius $r$ : $v_{0} = \frac{GM}{r} = \frac{g R ^{2}}{r}$
Time period: $T = 2 π \frac{r ^{3}}{GM} = 2 π \frac{( R + h ) ^{3}}{GM}$
Angular speed: $ω = \frac{GM}{r ^{3}}$
Kinetic energy: $K = \frac{1}{2} m v_{0}^{2} = \frac{GM m}{2 r}$
Potential energy: $U = - \frac{GM m}{r}$
Total energy: $E = K + U = - \frac{GM m}{2 r}$ (bound: $E < 0$ )
Binding energy: $B = ∣ E ∣ = \frac{GM m}{2 r}$
Escape velocity: $v_{e} = \frac{2 GM}{R} = 2 g R \approx 11.2 km/s$
Angular momentum: $L = m v_{0} r = m GM r$
Kepler's third law: $T^{2} = \frac{4 π ^{2}}{GM} a^{3} ⟹ T^{2} \propto a^{3}$
Areal velocity: $\frac{d A}{d t} = \frac{L}{2 m}$ (constant)

1. Motion of a Satellite in a Circular Orbit

Consider a satellite of mass $m$ in a circular orbit of radius $r$ around a planet of mass $M$ . The only significant force is the planet's gravity, and it must provide the centripetal force for circular motion:

$\frac{GM m}{r ^{2}} = \frac{m v _{0}^{2}}{r}$

Figure 1: Satellite of mass

m

in circular orbit of radius

r

. Gravity supplies the centripetal force:

\frac{GM m}{r ^{2}} = \frac{m v _{0}^{2}}{r}

, giving orbital speed

v_{0} = GM / r

Orbital speed

Solving for $v_{0}$ :

v_{0} = \frac{GM}{r}

Using $g = GM / R^{2}$ at the planet's surface, this can also be written:

$v_{0} = R \frac{g}{r} = \frac{g R ^{2}}{r}$

For a satellite at altitude $h$ above the surface, $r = R + h$ , so $v_{0} = g R^{2} / (R + h)$ .

Angular speed and time period

$ω = \frac{v _{0}}{r} = \frac{GM}{r ^{3}}$

$T = \frac{2 π}{ω} = 2 π \frac{r ^{3}}{GM} = 2 π \frac{( R + h ) ^{3}}{GM}$

Near-Earth satellite ( $h ≪ R$ ): Taking

r \approx R = 6400 km

and

g = 9.8 m/s^{2}

$v_{0} = g R \approx 9.8 \times 6.4 \times 1 0^{6} \approx 7.9 km/s$
$T = 2 π R / g \approx 84 minutes$

This is why low-orbit satellites (like the ISS, at about 400 km altitude) complete a revolution in roughly 90 minutes.

2. Energy of a Satellite in Orbit

Kinetic energy

Using $v_{0}^{2} = GM / r$ :

$K = \frac{1}{2} m v_{0}^{2} = \frac{GM m}{2 r}$

Potential energy

Taking $U = 0$ at infinity:

$U = - \frac{GM m}{r}$

Total mechanical energy

$E = K + U = \frac{GM m}{2 r} - \frac{GM m}{r} = - \frac{GM m}{2 r}$

E = - \frac{GM m}{2 r}

Three important relations follow:

$E = - K$ (total energy equals negative of kinetic energy)
$E = U /2$ (total energy equals half of potential energy)
$E < 0$ , indicating the satellite is bound to the planet.

Binding energy

The binding energy is the minimum energy that must be supplied from outside to just remove the satellite from orbit (send it to infinity with zero speed):

B = - E = \frac{GM m}{2 r}

For a satellite on Earth's surface ( $r = R$ ), the binding energy is $\frac{1}{2} m g R$ . This is exactly half of $m g R$ - the rest is supplied as kinetic energy at launch (see escape velocity, Section 5).

Angular momentum

For a circular orbit, $L$ is perpendicular to the orbital plane, with magnitude:

$L = m v_{0} r = m \frac{GM}{r} \cdot r = m GM r$

Since gravity is a central force, $L$ is conserved. For elliptical orbits this leads directly to Kepler's second law (Section 7).

3. Types of Satellites

Geostationary satellite

A geostationary satellite appears fixed above a point on the equator. For this:

Its time period must equal Earth's rotation period: $T = 24 hours = 86400 s$ .
It must rotate in the same sense as Earth (west to east).
Its orbital plane must coincide with the equatorial plane.

Applying $T = 2 π r^{3} / GM$ with $T = 86400 s$ and $GM = g R^{2}$ gives:

$r = (\frac{GM T ^{2}}{4 π ^{2}})^{1/3} \approx 4.22 \times 1 0^{7} m = 42, 200 km$

The altitude above the surface is therefore $h = r - R \approx 42200 - 6400 = 35, 800 km \approx 36, 000 km$ .

Orbital speed at this radius: $v_{0} = g R^{2} / r \approx 3.07 km/s$ .

Applications: Communications, direct-broadcast TV, weather (INSAT, GSAT series). Because the satellite appears fixed, ground antennas need no tracking.

Polar satellite

A polar satellite orbits in a plane that passes over (or close to) both geographic poles. Because Earth rotates beneath the satellite's fixed orbital plane, over successive orbits the satellite scans the entire surface - useful for remote sensing, weather, and reconnaissance.

Typical altitude: 500 - 1000 km (low orbit).
Typical period: about 100 minutes.
Indian examples: IRS, Cartosat, RISAT series.

4. Weightlessness in a Satellite

An astronaut inside an orbiting satellite floats freely and feels weightless, yet gravity at that altitude is nearly the same as at the surface (only slightly reduced). Why?

Both the satellite and everything inside it are in continuous free fall toward Earth's centre. Their common acceleration is:

$g^{'} = \frac{GM}{r ^{2}}$

The satellite's floor accelerates inward at exactly the same rate as the astronaut. So the floor exerts no normal contact force on the astronaut - and it is this normal force that we perceive as "weight" in everyday life.

Figure 2: Weightlessness in orbit. The satellite and everything inside share the same free-fall acceleration

g^{'} = GM / r^{2}

, so relative to the satellite floor the astronaut experiences no normal force. True weight

m g^{'}

is nonzero, but apparent weight is zero.

True weight vs apparent weight:

True weight $= m g^{'} \neq = 0$ (gravity is still pulling).
Apparent weight $= N = 0$ (no normal force from the floor).

The same weightless sensation is felt momentarily inside a freely falling lift, a projectile at the top of its arc, or an aircraft flying a parabolic "zero-g" trajectory.

5. Escape Velocity

The escape velocity is the minimum speed with which a body must be projected from the surface of a planet so that it just escapes the planet's gravitational field, reaching infinity with zero kinetic energy.

Using conservation of energy between the surface and infinity:

$\frac{1}{2} m v_{e}^{2} - \frac{GM m}{R} = 0 + 0$

v_{e} = \frac{2 GM}{R} = 2 g R

For Earth: $v_{e} = 2 \times 9.8 \times 6.4 \times 1 0^{6} \approx 11.2 km/s = 11, 200 m/s$ .

Key facts:

$v_{e}$ is independent of the mass of the projectile - a marble and a rocket have the same escape speed from Earth.
$v_{e}$ depends only on the mass and radius of the planet.
Relation to orbital speed: $v_{e} = 2 v_{0} \approx 1.414 v_{0}$ (for the same launch radius).
Escape from the Moon: $\approx 2.4 km/s$ . Escape from the Sun (from Earth's orbit): $\approx 42.1 km/s$ .

Solved Example 1

A body is projected vertically upward from the Earth's surface with a velocity just sufficient to carry it to infinity. Calculate the time taken to reach height

h

above the surface.

Solution:

Since the body just reaches infinity with zero speed, it was projected at escape speed $v_{e} = 2 g R$ . At distance $r$ from Earth's centre, energy conservation gives:

$\frac{1}{2} m v^{2} - \frac{GM m}{r} = \frac{1}{2} m v_{e}^{2} - \frac{GM m}{R}$

The right side is zero (escape condition). Using $GM = g R^{2}$ :

$v^{2} = \frac{2 g R ^{2}}{r}$

So $\frac{d r}{d t} = R \frac{2 g}{r}$ , giving:

$\int_{0}^{t} d t = \frac{1}{R 2 g} \int_{R}^{R + h} r^{1/2} d r$

Integrating:

$t = \frac{1}{R 2 g} \cdot \frac{2}{3} [(R + h)^{3/2} - R^{3/2}]$

$t = \frac{1}{3} \frac{2 R}{g} [(1 + \frac{h}{R})^{3/2} - 1]$

6. Bound and Unbound Trajectories

The shape of a satellite's trajectory near a planet depends on the launch speed $v$ compared with the circular orbital speed $V_{c} = GM / r$ and the escape speed $V_{e} = 2 GM / r$ at the launch radius:

Launch speed	Trajectory	Bound / Unbound
$v < V_{c}$	Ellipse with Earth's centre as the farther focus; the body hits the surface before completing the orbit	Bound (impact)
$v = V_{c}$	Circular orbit centred on the planet	Bound
$V_{c} < v < V_{e}$	Ellipse with Earth's centre as the nearer focus	Bound
$v = V_{e}$	Parabolic path - just escapes with zero speed at infinity	Marginal (unbound)
$v > V_{e}$	Hyperbolic path - escapes with nonzero speed at infinity	Unbound

Total energy determines the type:

E < 0

(ellipse/circle - bound),

E = 0

(parabola - marginal),

E > 0

(hyperbola - unbound). Angular momentum determines the shape within each type.

7. Kepler's Laws of Planetary Motion

Johannes Kepler (1571-1630) distilled Tycho Brahe's naked-eye planetary observations into three empirical laws. Newton later showed all three follow from his law of gravitation applied to a central inverse-square force.

First law (Law of orbits)

Every planet moves in an ellipse with the Sun at one focus. Circles are a special case of ellipses (with eccentricity $e = 0$ ).

Figure 3: Kepler's first law. Planet moves in an ellipse with the Sun at one focus. Perihelion distance

r_{p} = a (1 - e)

; aphelion distance

r_{a} = a (1 + e)

, where

a

is the semi-major axis and

e

the eccentricity.

For an ellipse of semi-major axis $a$ and eccentricity $e$ :

Perihelion distance (closest to Sun): $r_{p} = a (1 - e)$
Aphelion distance (farthest from Sun): $r_{a} = a (1 + e)$
Semi-minor axis: $b = a 1 - e^{2}$

Second law (Law of areas)

The line joining a planet to the Sun sweeps out equal areas in equal intervals of time. Equivalently, the areal velocity is constant:

\frac{d A}{d t} = \frac{L}{2 m} = constant

This is a direct consequence of the conservation of angular momentum $L$ , which itself follows from gravity being a central force ( $τ = r \times F = 0$ when $F$ is along $r$ ).

Figure 4: Law of areas. In equal time intervals, the radius vector from the Sun to the planet sweeps out equal areas. Areal velocity

\frac{d A}{d t} = \frac{L}{2 m}

is constant, since angular momentum

L

is conserved.

Derivation of $d A / d t = L /2 m$ : In a small time $d t$ , the radius vector sweeps a triangular area:

$d A = \frac{1}{2} ∣ r \times d r ∣ = \frac{1}{2} ∣ r \times v ∣ d t$

Since $L = m (r \times v)$ :

$\frac{d A}{d t} = \frac{∣ r \times v ∣}{2} = \frac{L}{2 m}$

Because $L$ is conserved, $d A / d t$ is a constant of motion - equal areas in equal times.

Third law (Law of periods)

The square of the time period of a planet is proportional to the cube of the semi-major axis of its orbit:

T^{2} \propto a^{3} or T^{2} = \frac{4 π ^{2}}{GM} a^{3}

For any two planets orbiting the same central body:

$(\frac{T _{1}}{T _{2}})^{2} = (\frac{a _{1}}{a _{2}})^{3}$

This law lets us compute one planet's period (or orbit size) from another's without knowing $GM$ - Kepler applied it purely empirically, long before Newton.

Solved Example 2

A body orbits the Earth at a mean radius

9

times that of a geostationary satellite. In how many days does it complete one revolution, and what is its angular velocity?

Solution:

Applying Kepler's third law with $R_{2} = 9 R_{1}$ and $T_{1} = 1$ day:

$(\frac{T _{2}}{T _{1}})^{2} = (\frac{R _{2}}{R _{1}})^{3} = 9^{3} = 729$

$T_{2} = 729 \times 1 day = 27 days$

Angular velocity:

$ω = \frac{2 π}{T _{2}} = \frac{2 π}{27 \times 24 \times 3600 s} \approx 2.69 \times 1 0^{- 6} rad/s$

Solved Example 3

A planet of mass

m

moves along an ellipse around the Sun (mass

M

) so that its maximum and minimum distances from the Sun are

R

(aphelion) and

r

(perihelion). Find the angular momentum of this planet about the Sun.

Solution:

By conservation of angular momentum, if $v_{1}$ is speed at aphelion and $v_{2}$ at perihelion (both perpendicular to the radius vector at these points):

$m v_{1} R = m v_{2} r ⟹ v_{1} R = v_{2} r \dots (i)$

By conservation of energy:

$- \frac{GM m}{R} + \frac{1}{2} m v_{1}^{2} = - \frac{GM m}{r} + \frac{1}{2} m v_{2}^{2}$

Rearranging:

$\frac{v _{2}^{2} - v _{1}^{2}}{2} = GM (\frac{1}{r} - \frac{1}{R}) = \frac{GM ( R - r )}{R r}$

From (i), $v_{2} = v_{1} R / r$ . Substituting:

$\frac{v _{1}^{2}}{2} (\frac{R ^{2}}{r ^{2}} - 1) = \frac{GM ( R - r )}{R r}$

$\frac{v _{1}^{2}}{2} \cdot \frac{R ^{2} - r ^{2}}{r ^{2}} = \frac{GM ( R - r )}{R r}$

Cancelling $(R - r)$ (using $R^{2} - r^{2} = (R - r) (R + r)$ ):

$v_{1}^{2} = \frac{2 GM r}{R ( R + r )}$

Therefore, angular momentum about the Sun is:

$L = m v_{1} R = m \frac{2 GM r R}{R + r}$

Common Mistakes to Avoid

Watch out

Confusing orbital speed with escape speed: $v_{0} = GM / r$ keeps a body in orbit; $v_{e} = 2 GM / r$ lets it just escape. Escape speed is always $2$ times orbital speed at the same radius.
Thinking astronauts feel weightless because gravity is absent: Gravity at the ISS altitude (~400 km) is still about $8.7 m/s^{2}$ - almost as strong as at the surface. Astronauts feel weightless because they are in free fall along with the satellite.
Using $T^{2} = k a^{3}$ across different central bodies: The constant $k = 4 π^{2} / GM$ depends on the mass $M$ of the body being orbited. You cannot compare a moon of Jupiter with a planet of the Sun using the same ratio.
Assuming geostationary means "stationary in space": Geostationary satellites are stationary relative to Earth's surface, but they still move at about 3 km/s in the inertial frame - matching Earth's rotation.
Forgetting the negative sign in total energy: For a bound orbit $E = - GM m /2 r$ ; the minus sign is essential. A "small" total energy for a satellite actually means a strongly negative number (tightly bound). Energy becomes less negative as the orbit gets larger.
Applying Kepler's second law only to elliptical orbits: It holds for any central force, including circular orbits (where $d A / d t$ is trivially constant because $r$ and $v$ are both constant).
Confusing perihelion/aphelion speeds: By angular momentum conservation, the planet is fastest at perihelion (closest to Sun) and slowest at aphelion (farthest). Beginners sometimes get this reversed.

Frequently Asked Questions

What is the orbital speed of a satellite?

The orbital speed of a satellite at distance $r$ from the centre of a planet of mass $M$ is $v_{0} = GM / r$ . For a low-Earth-orbit satellite (radius approximately equal to Earth's radius), this is about $7.9 km/s$ . Orbital speed decreases as orbit radius increases: $v_{0} \propto 1/ r$ .

What is escape velocity and what is its value for Earth?

Escape velocity is the minimum speed needed to project a body from a planet's surface so that it just escapes to infinity with zero kinetic energy. For Earth, $v_{e} = 2 g R \approx 11.2 km/s$ . It depends only on the planet's mass and radius, not on the mass of the escaping body.

State Kepler's three laws of planetary motion.

(1) Law of orbits: Every planet moves in an ellipse with the Sun at one focus. (2) Law of areas: The radius vector from Sun to planet sweeps out equal areas in equal times (constant areal velocity). (3) Law of periods: The square of a planet's time period is proportional to the cube of the semi-major axis of its orbit, $T^{2} \propto a^{3}$ .

What is a geostationary satellite? At what altitude does it orbit?

A geostationary satellite has the same orbital period as Earth's rotation (24 hours) and orbits in the equatorial plane in the same sense as Earth's rotation, so it appears fixed above one point on the equator. Its orbit radius is about $42, 200 km$ from Earth's centre, giving an altitude of approximately $36, 000 km$ above the surface.

What is a polar satellite and how is it different from a geostationary satellite?

A polar satellite orbits in a plane that passes over (or near) both geographic poles, at low altitude (typically 500 - 1000 km) with a period of about 100 minutes. As Earth rotates beneath the fixed orbital plane, the satellite scans the entire surface over successive orbits - useful for remote sensing and weather monitoring. In contrast, geostationary satellites stay above one point on the equator at 36,000 km altitude.

Why do astronauts feel weightless inside an orbiting satellite?

An orbiting satellite is in continuous free fall toward Earth; the astronaut inside falls at the same rate. Since the satellite's floor accelerates inward at exactly the same rate as the astronaut, there is no normal contact force between them - and it is this normal force we perceive as \"weight\". True weight (gravitational pull) is still nearly full strength; only the apparent weight is zero.

What is the total energy of a satellite in orbit and why is it negative?

For a satellite of mass $m$ in a circular orbit of radius $r$ around a planet of mass $M$ , the total mechanical energy is $E = K + U = - GM m /2 r$ . The negative sign means the satellite is bound - external energy equal to $∣ E ∣$ (the binding energy) must be supplied to remove it to infinity. A positive total energy would mean an unbound (hyperbolic) trajectory.

What is the relation between escape velocity and orbital velocity?

At the same radius $r$ , escape velocity $v_{e} = 2 GM / r$ and orbital velocity $v_{0} = GM / r$ . Therefore $v_{e} = 2 v_{0} \approx 1.414 v_{0}$ . In terms of energy, doubling the kinetic energy of a satellite (multiplying its speed by $2$ ) is exactly what it takes to unbind it from orbit.

What is areal velocity and why is it constant for a planet?

Areal velocity is the rate at which the radius vector from the Sun to the planet sweeps out area, $d A / d t = L / (2 m)$ , where $L$ is the planet's angular momentum. Because gravity is a central force (always along the radius vector), it exerts no torque about the Sun, so $L$ is conserved - and hence $d A / d t$ is constant. This is precisely Kepler's second law.

Previous year questions on Satellite, Planetary Motion And Kepler’s Law

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