Gravitation: common doubts, answered
The questions students ask most often about Gravitation, each with a short answer. For the full chapter, read the Gravitation notes.
Kepler's laws
Read this section in the notes →What are Kepler's three laws of planetary motion?
First, every planet follows an elliptical orbit, and the Sun sits at one of the two foci. Second, the area traced by the Sun-planet line grows at a steady rate, so equal times cover equal areas. Third, the orbital period squared is proportional to the semi-major axis cubed. Kepler reached these from careful observations before Newton explained them with gravity.
Why does the period of a planet depend on its distance from the Sun?
Gravity weakens with distance, so a farther planet moves more slowly and also has a longer path to cover. Combining the two gives T² = (4π² / G M) a³, so the period grows as the 3/2 power of the orbit size. The planet's own mass does not appear, so the relation holds for every planet around the same Sun.
The law of areas
Read this section in the notes →Why does a planet move faster when it is closer to the Sun?
Because the line to the planet must sweep equal areas in equal times. Near the Sun the line is short, so the planet has to move faster to cover the same area; far away the line is long and the planet moves slowly. Physically, this comes from conservation of angular momentum, since gravity points towards the Sun and gives no torque about it.
Does the law of areas prove the inverse-square law of gravity?
No. The law of areas follows from angular momentum being conserved, which holds for any central force, one directed along the line to a fixed centre, whatever its dependence on distance. It shows that gravity is central, but not that it falls as 1/r². The inverse-square form is linked to Kepler's third law instead.
Universal law of gravitation
Read this section in the notes →What should r be in Newton's law of gravitation?
r is the distance between the centres of the two bodies, not the gap between their surfaces. For a satellite at height h above the Earth, use r = R_E + h in F = G m₁ m₂ / r². Treating a uniform sphere as if all its mass were at its centre is valid for points outside the sphere.
Why don't we feel the gravitational pull between everyday objects?
Because G is extremely small, about 6.67 × 10⁻¹¹ N m² kg⁻², so the force between two ordinary objects is tiny. Two people a metre apart attract each other with a force far too small to notice against friction and other forces. Gravity becomes important only when at least one body is enormous, like the Earth.
The gravitational constant
Read this section in the notes →What is the difference between G and g?
G is the universal gravitational constant, the same everywhere in the universe. g is the acceleration due to gravity at a place, which depends on the planet's mass and radius and on height or depth. On the Earth's surface g = G M_E / R_E², about 9.8 m s⁻², while G is about 6.67 × 10⁻¹¹ N m² kg⁻².
How was G measured?
Cavendish measured it with a torsion balance: small lead spheres on a light rod hung from a fine wire, with large spheres placed near them. The gravitational attraction twisted the wire by a small angle, and the restoring torque of the wire balanced it. Knowing the masses, distances and twist gave the value of G.
Acceleration due to gravity
Read this section in the notes →Why is g the same for all bodies at a place?
Because the gravitational force on a body is proportional to its mass, and Newton's second law divides that force by the same mass. The mass cancels, leaving g = G M_E / R_E² for any body near the surface. So a heavy stone and a light one fall with equal acceleration when air resistance is negligible.
g above and below the surface
Read this section in the notes →How does g change with height above the Earth?
It decreases, following g(h) = g R_E² / (R_E + h)². For heights much smaller than the Earth's radius, this is approximately g(1 − 2h/R_E). The approximate form is valid only when h is small compared with R_E; for a height like R_E/2, use the exact expression instead.
Does g increase as you go deeper into the Earth?
No, for a uniform Earth it decreases: g(d) = g(1 − d/R_E). Going down brings you closer to the centre, but the shell of Earth above you exerts no net force, so less mass pulls you. g becomes zero at the centre. So g is greatest at the surface and falls both above and below it.
Gravitational potential energy
Read this section in the notes →Why is gravitational potential energy negative?
Because the zero is chosen at infinite separation, and gravity is attractive. To pull two bodies apart to infinity you must do positive work, so their energy when close must be less than zero: U = −G m₁ m₂ / r. The negative sign means the bodies are bound; only differences in potential energy have physical meaning.
When can you use mgh for gravitational potential energy?
Only for heights small compared with the Earth's radius, where g is nearly constant. Then the change in potential energy for a rise h is about mgh. For satellites or large heights, use the full form U = −G M m / r and take the difference between the two values of r.
Escape speed
Read this section in the notes →Does escape speed depend on the mass of the object?
No. The escape speed vₑ = √(2 G M / R) depends only on the planet's mass and radius, not on the mass of the escaping body. A heavier rocket needs more energy to escape, but the same speed, because both its kinetic energy and the energy needed scale with its mass. For the Earth vₑ is about 11.2 km/s.
Why does the Moon have no atmosphere?
The Moon's escape speed is only about 2.3 km/s, much lower than the Earth's, because its mass is small. Gas molecules there, heated by sunlight, can reach speeds high enough to escape. Over time almost all gas has leaked away, leaving the Moon with no atmosphere to speak of.
Earth satellites
Read this section in the notes →Does the orbital speed of a satellite depend on its mass?
No. For a circular orbit, gravity provides the centripetal force, and the satellite's mass cancels, giving v = √(G M_E / (R_E + h)). Its speed depends only on the Earth's mass and the orbit radius. Higher orbits have lower speeds and longer periods.
Why do astronauts feel weightless in orbit?
Not because gravity is absent; at typical orbit heights g is still most of its surface value. The astronaut and the spacecraft are both in free fall around the Earth with the same acceleration, so the floor does not push on the astronaut. With no normal force, there is no sensation of weight.
What is the period of a satellite orbiting just above the Earth's surface?
About 85 minutes. With the radius close to R_E, the period becomes T₀ = 2π √(R_E / g). Putting R_E ≈ 6.4 × 10⁶ m and g ≈ 9.8 m s⁻² gives roughly 5100 s. The corresponding orbital speed is v₀ = √(g R_E), close to 8 km/s.
Energy of an orbiting satellite
Read this section in the notes →Why is the total energy of an orbiting satellite negative?
Because its potential energy, −G M m / r, is twice as large in size as its kinetic energy, G M m / 2r. Adding them gives E = −G M m / 2r, a negative value that shows the satellite is bound to the Earth. If its total energy reached zero, it would escape.
How much energy is needed to move a satellite to a higher orbit?
The energy needed is the difference in total energy between the two orbits, E₂ − E₁, using E = −G M m / 2r for each. The higher orbit has a less negative total energy, so energy must be supplied. Interestingly, the satellite ends up with lower kinetic energy but much higher potential energy.
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