Summary — What This Chapter Covers
- Gravitational field strength g = F/m, in N kg−1; field lines show the direction of the force on a mass.
- Newton's law: F = Gm1m2/r2; outside a uniform sphere, treat its mass as a point at its centre.
- For a circular orbit, gravity provides the centripetal force: GMm/r2 = mv2/r.
- Geostationary orbit: period 24 h, west to east, directly above the Equator.
- g = GM/r2; potential ϕ = −GM/r; potential energy EP = −GMm/r.
1. Gravitational Field
13.1
A field of force
A field of force is a region in which an object experiences a force. A gravitational field is a region in which a mass experiences a force. The gravitational field strength at a point is the force per unit mass on a small mass placed there. It is a vector, and its unit N kg−1 is equivalent to m s−2.
Field lines show the direction of the force on a mass. Around a spherical mass they are radial and point inwards; the closer the lines, the stronger the field. Near the Earth’s surface, over a small region, the lines are parallel and equally spaced: the field is uniform.
Definition
g = F/m
Worked Example
A 2.0 kg mass on the Moon’s surface has a weight of 3.2 N. Find the Moon’s surface field strength.
2. Gravitational Force Between Point Masses
13.2
Newton's law of gravitation
Two point masses attract each other with a force proportional to the product of their masses and inversely proportional to the square of their separation. For a point outside a uniform sphere, the sphere behaves as if all its mass were concentrated at its centre, so r is measured centre to centre.
Orbits. For a satellite in a circular orbit, gravity is the only force and provides the centripetal force. Equating gives v2 = GM/r and, using v = 2πr/T, T2 = (4π2/GM) r3. The satellite’s own mass cancels.
A geostationary satellite stays above the same point on the Earth: period 24 hours, orbiting west to east (the direction the Earth spins), in the plane of the Equator.
Newton's law and orbits
F = Gm1m2/r2 GMm/r2 = mv2/r = mrω2
Worked Example
The Earth has mass 5.97 × 1024 kg. Find the radius of a geostationary orbit.
Practice Question
Find the gravitational force between the Earth (5.97 × 1024 kg) and the Moon (7.35 × 1022 kg), 3.84 × 108 m apart.
3. Gravitational Field of a Point Mass
13.3
g = GM/r²
Put a small test mass m at distance r from a point mass M. The force on it is GMm/r2, and dividing by m gives the field strength. So g falls off with the inverse square of distance from the centre.
Near the Earth’s surface a change in height of a few kilometres changes r by a tiny fraction of the Earth’s radius (6.37 × 106 m), so g is approximately constant. That is why we can use g = 9.81 m s−2 for everyday heights.
Derived result
g = F/m = (GMm/r2)/m = GM/r2
Worked Example
Taking M = 5.97 × 1024 kg and R = 6.37 × 106 m, find g at the Earth’s surface and at a height of 400 km.
Memorise