The motion of a satellite around a central body is governed by universal gravitation acting as a centripetal force. It describes the orbital mechanics of natural and artificial satellites under conservative central force fields.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Orbital velocity, time period of satellites, height of geostationary satellites, and total energy of a satellite in circular orbit.
PGT extension
Elliptical orbits, areal velocity conservation (Kepler's second law), angular momentum conservation, escape velocity, and orbital perturbations.
Prerequisites
Newton's Law of Gravitation
Centripetal Force and Acceleration
Conservation of Angular Momentum
Orbital Velocity and Height
The orbital velocity of a satellite depends on the mass of the central body and the radius of the orbit, but is independent of the mass of the satellite.
For a satellite revolving close to the Earth's surface, the orbital velocity is approximately 7.9 km/s.
As the orbital radius increases, the orbital velocity decreases while the potential energy increases.
Time Period and Geostationary Orbit
The square of the time period of a satellite is directly proportional to the cube of its orbital radius (Kepler's Third Law).
A geostationary satellite rotates in the equatorial plane with a time period of exactly 24 hours, matching the Earth's rotational period.
The height of a geostationary satellite is approximately 35,800 km above the Earth's surface.
Energy of an Orbiting Satellite
The total mechanical energy of a bound satellite is negative, indicating a bound system under an attractive central force.
Kinetic energy is equal in magnitude to half of the potential energy and negative of the total energy.
Binding energy is defined as the negative of the total mechanical energy required to remove the satellite to infinity.
Kepler's Laws of Planetary Motion
First Law (Law of Orbits): Every satellite moves in an elliptical orbit with the central body at one of the foci.
Second Law (Law of Areas): The position vector of the satellite sweeps out equal areas in equal intervals of time, reflecting conservation of angular momentum.
Third Law (Law of Periods): The square of the time period is proportional to the cube of the semi-major axis.
Derivations
Orbital Velocity and Time Period of a Satellite
Conditions: Circular orbit of radius $r = R + h around a spherical Earth of mass M$ and radius R.
Equate the gravitational force to the required centripetal force: \frac{GMm}{$r^2 = \frac{mv_0^2}{r}$
Solve for orbital velocity: $v_0 = \sqrt{\frac{GM}{r}} = \sqrt{\frac{GM}{R+h}}$
Express time period T as distance over orbital speed: $T = \frac{2\pi r}{v_0} = 2\pi \sqrt{\frac{r^3}{GM}}$
Result: $v_0 = \sqrt{\frac{GM}{R+h}}$ and $T = 2\pi \sqrt{\frac{(R+h)^3}{GM}}$
Total Energy of a Satellite in Circular Orbit
Conditions: Satellite of mass m in a circular orbit of radius r around mass M.
Write kinetic energy from centripetal condition \frac{mv^2}{r} = \frac{GMm}{r^2}: $K = \frac{1}{2}mv^2 = \frac{GMm}{2r}$
Sum kinetic and potential energy for total mechanical energy E: $E = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$
Result: $E = -\frac{GMm}{2r}$
Formula sheet
Formula / law
Equation
Variables and units
Conditions
Orbital Velocity
$$v_0 = \sqrt{\frac{GM}{R+h}}$$
G = Universal gravitational constant, M = Mass of Earth, R = Radius of Earth, h = Height above surface m/s
Circular orbit
Time Period
$$T = 2\pi \sqrt{\frac{(R+h)^3}{GM}}$$
R = Radius of Earth, h = Height, G = Gravitational constant, M = Mass of Earth s
Circular orbit
Gravitational Acceleration at Height h
$$g_h = g\left(1 - \frac{2h}{R}\right)$$
g = Surface gravity, h = Height (h << R), R = Radius of Earth m/s^2
h is very small compared to R
Kinetic Energy
$$K = \frac{GMm}{2r}$$
G = Gravitational constant, M = Mass of central body, m = Mass of satellite, r = Orbital radius J
Circular orbit
Potential Energy
$$U = -\frac{GMm}{r}$$
G, M, m, r with zero potential at infinity J
Standard gravitational reference
Total Mechanical Energy
$$E = -\frac{GMm}{2r}$$
G, M, m, r J
Bound circular orbit
Angular Momentum
$$L = m \sqrt{GMr}$$
m = Mass of satellite, G, M, r = Orbital radius kg m^2/s
Circular orbit
Escape Velocity
$$v_e = \sqrt{2gR} = \sqrt{\frac{2GM}{R}}$$
g = Acceleration due to gravity, R = Radius of Earth m/s
Fired from surface of Earth to infinity
Graphs and experiments
Potential, Kinetic, and Total Energy vs. Orbital Radius
Draw: Plot of U (negative hyperbola), K (positive hyperbola), and E (negative hyperbola below U) as a function of r.
Interpret: Shows that total energy is negative (bound state) and kinetic energy is half the magnitude of potential energy.
Exam use: Direct identification of energy curves and bound system conditions in multiple-choice questions.
Square of Time Period vs. Cube of Orbital Radius (T^2 vs. r^3)
Draw: A straight line passing through the origin with slope 4π^2 / (GM).
Interpret: Verifies Kepler's Third Law, confirming the inverse-square nature of gravitational force.
Exam use: Interpreting slope values to calculate central mass M.
Orbital Velocity vs. Orbital Radius
Draw: A decaying curve representing inverse square root dependence v proportional to 1 / sqrt(r).
Interpret: Demonstrates that satellites closer to the central body travel faster than those farther away.
Exam use: Conceptual questions comparing speeds of Low Earth Orbit (LEO) and Geostationary satellites.
Solved numericals
TGT core
Calculate the orbital velocity of an artificial satellite revolving in a circular orbit at a height of 600 km above the Earth's surface. (Take R = 6400 km, $g = 9.8 m/s^2)$
Identify given parameters: $R = 6400 km = 6.4 \times 10^6 m$, $h = 600 km = 0.6 \times 10^6 m$, $g = 9.8 m/s^2$.
Calculate orbital radius $r = R + h = 6.4 \times 10^6 + 0.6 \times 10^6 = 7.0 \times 10^6 m$.
Use formula $v_0 = \sqrt{gR^2 / r} or v_0 = \sqrt{GM / r}$ where G$M = gR^2$.
A satellite of mass 200 kg is revolving in a circular orbit of radius 2R around the Earth. Find its kinetic energy, potential energy, and total energy. (Take R = 6400 km, $g = 9.8 m/s^2)$
Identify given parameters: m = 200 kg, $r = 2R = 2 \times 6.4 \times 10^6 m = 1.28 \times 10^7 m$, G$M = gR^2$.
Calculate Potential Energy $U = -GMm / r = -gR^2 m / (2R) = -gRm / 2$.
Calculate Total Energy $E = K + U = -3.136 \times 10^9 J$.
Answer: $K = 3.14 x 10^9 J$, $U = -6.27 x 10^9 J$, $E = -3.14 x 10^9 J$
PGT extension
Find the minimum energy required to launch a satellite of mass 1000 kg from the Earth's surface into a circular orbit at an altitude equal to the Earth's radius R.
Initial energy of the satellite on Earth's surface: $E_i = K_i + U_i = 0 - \frac{GMm}{R} = -\frac{GMm}{R}$.
Final total energy in circular orbit at $r = R + R = 2R: E_f = -\frac{GMm}{2(2R)} = -\frac{GMm}{4R}$.
The central body is assumed to be spherically symmetric with a uniform mass distribution.
Atmospheric drag and dissipative frictional forces are neglected unless explicitly stated.
Gravitational perturbations from the Sun, Moon, and non-spherical Earth oblateness (J2 term) are ignored in basic circular orbit models.
Common misconceptions
Common error: A satellite in orbit is beyond the Earth's gravitational pull, which is why astronauts float.
Correct: Gravity is very much present; astronauts float because both the satellite and the astronauts are in free fall together towards the Earth.
Centripetal acceleration equals local acceleration due to gravity, producing apparent weightlessness.
Common error: A heavier satellite requires a higher orbital velocity to stay in the same orbit.
Correct: Orbital velocity is independent of the mass of the satellite.
Mass cancels out when equating gravitational force and centripetal force.
Common error: Total energy of a satellite can be positive.
Correct: Total energy of a bound satellite is always negative.
A positive total energy implies an unbound hyperbolic or parabolic trajectory, not a closed orbit.
Quick revision
Orbital velocity varies inversely as the square root of the orbital radius.
Orbital Velocity: $v_0 = \sqrt{\frac{GM}{R+h}}$
Time Period: $T = 2\pi \sqrt{\frac{(R+h)^3}{GM}}$
Must remember
Orbital velocity of a satellite is independent of its own mass and decreases as the orbital radius increases ($v_0 = \sqrt{frac{GM}{r}}$).
The total mechanical energy of a bound orbiting satellite is always negative ($E = -\frac{GMm}{2r}$).
For a geostationary satellite, the time period is exactly $24\text{ hours}$ and its height above the Earth's surface is approximately $35,800\text{ km}$.
Kepler's Second Law (Law of Areas) is a direct consequence of the conservation of angular momentum for a central force.
The square of the time period of a satellite is directly proportional to the cube of its orbital radius or semi-major axis ($T^2 \propto r^3$).
Kinetic energy is equal in magnitude to half of the potential energy and negative of the total energy ($K = -E = |U|/2$).
Glossary
Orbital VelocityThe velocity required to keep a satellite in a stable circular orbit around a central body, given by $v_0 = \sqrt{\frac{GM}{r}}$.
Geostationary SatelliteA satellite revolving in the equatorial plane with a time period of $24\text{ hours}$, appearing stationary relative to the Earth's surface.
Binding EnergyThe negative of the total mechanical energy of a bound system, representing the minimum energy required to remove the satellite to infinity.
Escape VelocityThe minimum velocity required for an object to escape the gravitational field of a celestial body, given by $v_e = \sqrt{2gR}$.
Areal VelocityThe area swept out per unit time by the position vector of a revolving satellite, which remains constant due to conservation of angular momentum.
Orbital Velocity vs Escape Velocity
Parameter
Orbital Velocity ($v_0$)
Escape Velocity ($v_e$)
Expression (Surface)
$v_0 = \sqrt{gR}$
$v_e = \sqrt{2gR}$
Trajectory
Closed circular orbit
Parabolic or hyperbolic trajectory
Relation
$v_e = \sqrt{2} v_0$
$v_0 = \frac{v_e}{\sqrt{2}}$
Energy Parameters of a Satellite
Energy Type
Formula
Sign / Nature
Kinetic Energy ($K$)
$K = \frac{GMm}{2r}$
Always positive
Potential Energy ($U$)
$U = -\frac{GMm}{r}$
Always negative
Total Energy ($E$)
$E = -\frac{GMm}{2r}$
Always negative for bound orbit
Practice Quiz
Choose an option to reveal the explanation.
1. What is the orbital velocity of a satellite revolving close to the Earth's surface?
Explanation: For a satellite revolving very close to the Earth's surface ($h \approx 0$), the orbital velocity is approximately $7.9\text{ km/s}$, calculated using $v_0 = \sqrt{gR}$.
2. How does the orbital velocity of a satellite change as its orbital radius increases?
Explanation: Since orbital velocity is given by $v_0 = \sqrt{\frac{GM}{r}}$, it is inversely proportional to the square root of the orbital radius $r$. As $r$ increases, $v_0$ decreases.
3. Which physical quantity determines whether a satellite's orbit is bound or unbound?
Explanation: The total mechanical energy $E$ determines the boundedness. If $E < 0$, the orbit is bound (elliptic/circular); if $E \ge 0$, it is unbound (parabolic/hyperbolic).
4. What is the approximate height of a geostationary satellite above the Earth's surface?
Explanation: Using Kepler's Third Law with a time period of $24\text{ hours}$, the height of a geostationary satellite above the Earth's surface is calculated to be approximately $35,800\text{ km}$.
5. The total mechanical energy of an orbiting satellite in a circular orbit of radius $r$ is:
Explanation: The total energy is the sum of kinetic energy ($K = \frac{GMm}{2r}$) and potential energy ($U = -\frac{GMm}{r}$), yielding $E = -\frac{GMm}{2r}$.
6. Kepler's Second Law (Law of Areas) is a direct consequence of the conservation of:
Explanation: The areal velocity swept out by a satellite's position vector is constant, which is mathematically equivalent to the conservation of angular momentum under a central gravitational force.
7. Why do astronauts float inside an orbiting space station?
Explanation: Gravity is very much present at orbital heights. Astronauts float because both they and the spacecraft experience identical gravitational acceleration, resulting in free fall and apparent weightlessness.
8. If the mass of a satellite in a circular orbit is doubled while keeping the orbital radius constant, its orbital velocity will:
Explanation: Orbital velocity $v_0 = \sqrt{\frac{GM}{r}}$ depends only on the mass of the central body $M$ and the radius $r$; it is completely independent of the mass of the satellite $m$.
9. What is the ratio of kinetic energy to potential energy for a satellite in a circular orbit?
Explanation: Kinetic energy is $K = \frac{GMm}{2r}$ and potential energy is $U = -\frac{GMm}{r}$. Therefore, the ratio $K/U = -1/2$, or $K = -U/2$.
10. According to Kepler's Third Law, the square of the time period of a planet or satellite is proportional to:
Explanation: Kepler's Third Law states that $T^2 \propto r^3$, where $r$ is the orbital radius or semi-major axis of the orbit.
11. What is the binding energy of a satellite of mass $m$ orbiting at radius $r$ around a central body of mass $M$?
Explanation: Binding energy is defined as the negative of the total mechanical energy. Since $E = -\frac{GMm}{2r}$, the binding energy is $\frac{GMm}{2r}$.
12. A satellite moves in an elliptical orbit around the Sun. Where is its speed maximum?
Explanation: By conservation of angular momentum ($L = mvr = \text{constant}$), as the distance $r$ from the central body is minimum at the perihelion, the linear speed $v$ must be maximum.
13. What is the relation between escape velocity ($v_e$) and orbital velocity ($v_0$) for a satellite close to the Earth's surface?
Explanation: Orbital velocity near the surface is $v_0 = \sqrt{gR}$ and escape velocity is $v_e = \sqrt{2gR}$, which implies $v_e = \sqrt{2}v_0$.
14. If a satellite is given an additional boost such that its total mechanical energy becomes zero, its new trajectory will be:
Explanation: When total energy $E = 0$, the satellite escapes the gravitational bound and follows a parabolic trajectory.
15. Which of the following parameters does NOT affect the time period of a satellite orbiting in a circular path?
Explanation: The time period formula $T = 2\pi \sqrt{\frac{r^3}{GM}}$ depends on $r$, $G$, and $M$, but is independent of the mass of the satellite.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. If the radius of the orbit of a geostationary satellite is $4$ times that of a communication satellite revolving closer to Earth, what is the time period of the communication satellite?
Explanation: Using Kepler's Third Law $T^2 \propto r^3$, let $r_1$ and $T_1$ be for geostationary ($r_1 = 4r_2$, $T_1 = 24\text{ h}$). Then $(T_1/T_2)^2 = (r_1/r_2)^3 = (4)^3 = 64$. Thus $T_1/T_2 = 8$, giving $T_2 = 24/8 = 3\text{ hours}$ custodially.
2. A satellite of mass $m$ revolves around the Earth of mass $M$ in a circular orbit of radius $r$. The angular momentum of the satellite is given by:
Explanation: Angular momentum $L = m v_0 r$. Substituting $v_0 = \sqrt{\frac{GM}{r}}$, we get $L = m \sqrt{\frac{GM}{r}} r = m \sqrt{GMr}$.
3. If the gravitational acceleration $g$ suddenly becomes zero, a satellite currently in a circular orbit around the Earth will:
Explanation: Gravitational force provides the necessary centripetal force. If $g$ becomes zero, the centripetal force vanishes, and by Newton's first law, the satellite continues moving tangentially in a straight line.
4. The potential energy of a satellite of mass $m$ at a height equal to the Earth's radius $R$ above the surface is:
Explanation: The orbital radius at height $h = R$ is $r = R + h = R + R = 2R$. Potential energy $U = -\frac{GMm}{r} = -\frac{GMm}{2R}$.
5. The escape velocity from the surface of the Earth is $11.2\text{ km/s}$. What will be the escape velocity from a planet whose mass is twice and radius is half that of Earth?
Explanation: Escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$. For the new planet, $M' = 2M$ and $R' = R/2$, so $v_e' = \sqrt{\frac{2G(2M)}{R/2}} = 2\sqrt{\frac{2GM}{R}} = 2v_e = 2 \times 11.2 = 22.4\text{ km/s}$.
6. A satellite is orbiting the Earth in a circular orbit of radius $r$ with speed $v_0$. To make it escape from the Earth's gravitational field, its speed must be increased by a factor of:
Explanation: Escape velocity is $v_e = \sqrt{2}v_0$. The required increase in speed is $\Delta v = v_e - v_0 = (\sqrt{2} - 1)v_0$, which means it must be increased by a factor of $(\sqrt{2} - 1)$.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. What happens to the potential energy of a satellite when it moves from a lower circular orbit to a higher circular orbit?
Explanation: Potential energy is given by $U = -\frac{GMm}{r}$. As $r$ increases, the negative value decreases in magnitude, meaning the overall potential energy increases (becomes less negative).
2. What is the time period of a satellite revolving very close to the Earth's surface?
3. If a satellite is orbiting in a circular orbit of radius $r$ with angular velocity $\omega$, then according to Kepler's Third Law:
Explanation: Since $T = \frac{2\pi}{\omega}$, Kepler's Law $T^2 \propto r^3$ implies $\left(\frac{2\pi}{\omega}\right)^2 \propto r^3$, leading to $\omega^2 r^3 = \text{GM} = \text{constant}$.
4. The binding energy of a satellite is $E_B$. Its total mechanical energy $E$ is given by:
Explanation: Binding energy is defined as $E_B = -E$, which means total mechanical energy $E = -E_B$.
5. A planet moves around the Sun in an elliptical orbit. When is its kinetic energy maximum?
Explanation: Kinetic energy $K = \frac{1}{2}mv^2$. Speed $v$ is maximum at the perihelion (closest distance), hence kinetic energy is also maximum there.
6. The orbital velocity of a satellite at height $h$ above the Earth's surface when $h \ll R$ is proportional to:
Explanation: Orbital velocity is $v_0 = \sqrt{\frac{GM}{R+h}}$, which is proportional to $\frac{1}{\sqrt{R+h}}$.
7. What percentage change occurs in the time period of a satellite if its orbital radius is increased by $1\%$?
8. Which of the following statements is correct regarding the angular momentum vector $\vec{L}$ of a planet or satellite orbiting under central forces?
Explanation: Since torque due to central gravitational forces is zero (\vec{\tau} = \vec{r} \times \vec{F} = 0$), the angular momentum vector $\vec{L}$ remains strictly constant in magnitude and direction.
9. What is the work done by gravitational force on a satellite moving in a circular orbit of radius $r$ for one complete revolution?
Explanation: In a circular orbit, the gravitational force is always perpendicular to the instantaneous displacement of the satellite. Thus, the work done in one complete revolution is zero.
10. If the Earth were to suddenly shrink to half its present radius without any change in its mass, the acceleration due to gravity on its surface and the orbital velocity of a close satellite would:
Explanation: Surface gravity $g = \frac{GM}{R^2}$. If $R$ becomes $R/2$, $g$ increases by a factor of $4$. Similarly, close orbital velocity $v_0 = \sqrt{gR} = \sqrt{\frac{GM}{R}}$. If $R$ becomes $R/2$, $v_0$ increases by a factor of $\sqrt{2}$. Wait, let's re-verify: $v_0' = \sqrt{\frac{GM}{R/2}} = \sqrt{2} v_0$. Ah, let's check the options: $g$ increases by $4$ and $v_0$ increases by $\sqrt{2}$. Let's ensure option accuracy or pick the most rigorous one. Since options combine them, let's check option validity: wait, $g$ quadruples and $v_0$ increases by $\sqrt{2}$ is not listed together cleanly except option 2 or 3. Let's look at standard question: if $R$ is halved, $g$ becomes $4g$ and $v_0$ becomes $\sqrt{2}v_0$. Let's select option corresponding to correct physics.