No subscript is needed on the Fs, since the force is the same on both masses in magnitude, but opposite in direction (recall Newton’s third law). According to Newton's law of gravitation:

G is called the ‘universal gravitational constant’. By experiment, G = 6.67 * 10-11 N m2 kg-2

For objects of a finite size, r is the distance between their centres of mass. In particular, if the objects are spheres of uniform density, they can be treated as point objects with all their mass concentrated at their centres, so r is the distance between their centres:

G and g

Consider a mass m on the Earth’s surface:

The force on mass m equals its weight, mg. But this is due to gravity, so:


The radius of the Earth is 6.37 * 106m. What is its mass?   (G = 6.67*10-11 Nm2kg-2 ; g = 10ms-2)

GRAVITATIONAL FIELD STRENGTH (return to start of page)

i.e. if a mass m (kg) experiences a gravitational force F (newtons) then:

Thus, gravitational field strength and acceleration due to gravity are equivalent quantities.

Near the Earth’s surface:

So, as far calculations are concerned, numerically g = 10, whichever units it is actually given in.

Variation of g

a) Variation with height

In the above diagram, a>r, i.e. we consider a mass m at a point above the Earth, at which point:

b) Variation with depth

Making the (rough) assumption that the Earth has a uniform density, it can be shown that the acceleration due to gravity g/ at a point P below its surface is entirely due to the mass M/ within the sphere of radius a.

For a mass m at point P:

Variation of gravitational field strength (or acceleration due to gravity), g/ with depth below the Earth’s surface and height above it:

CIRCULAR ORBITS (return to start of page)

a) Velocity, v

At any instant, the mass m is moving at a tangent to the orbit and \ at 900 to radius R.

A centripetal force (which has a magnitude = mv2/R)  is needed to keep an object moving on a circular path, and in the present case, this is provided by the gravitational force:

b) Period, T

Close orbits

Suppose that a satellite is less than, say, 200km above the Earth, then, R is about equal to r (as 6600km is only a few percent bigger than the radius of the Earth, which is 6400km).

a) Velocity

b) Period

From (4):


Given g = 9.8ms-2  and the radius of Earth, r = 6.4*106 m, find the velocity and period for an object in an orbit close to the Earth.

Parking orbits

Suppose that the period of a satellite is exactly 24 hours, and that it is rotating above the equator and with the Earth, then the satellite will stay over the same place over the Earth’s surface. The satellite is said to be in a ‘parking orbit’ or a 'geostationary orbit' or a 'geosynchronous orbit'. Such satellites are used for communication and for observing the Earth.


For a satellite in a parking orbit, calculate:

  1. the radius of orbit
  2. the height above the Earth
  3. the speed in this orbit
(radius of the Earth = 6.4*106, mass of the Earth = 6.0*1024 kg, G = 6.7*10-11N m2 kg-2)

a) T = 24 hours = 24 * 60 * 60 seconds = 8.64 * 104s

b) height above the Earth = 42400 - 6400 = 36000 km


Suppose that an astronaut is in an orbit at a constant speed v, a distance R from the centre of the Earth. The astronaut and the spaceship undergo the same centripetal acceleration (= v2/R). Hence the floor of the spaceship ‘falls’ away from the astronaut at the same rate as s/he falls towards it. Hence the experience of ‘weightlessness’.

ESCAPE VELOCITY (return to start of page)

The total work done W in moving the mass m from r1 to r2 equals the area between the graph and the section of the axis from r1 to r2, and can be shown to be given by:

If the mass m is to be projected from the Earth’s surface (r1 = rE ) and is to escape to infinity (r2 = infinity), then:

To escape to infinity from the Earth’s surface, the body must projected with this amount of initial energy as kinetic energy.

Notice that the mass of the object does not appear in the equation.


Determine the escape velocity from the Earth given that:

Note: The above equation only applies to projectiles. A body in powered flight does not have to rely upon its initial KE to escape from the Earth, so does not need reach the escape velocity (given a long enough ladder you could walk to the moon!)

GRAVITATIONAL POTENTIAL AT A POINT (return to start of page)


Potential is a scalar quantity.

From earlier, the work done in moving a mass m from the Earth’s surface to infinity is given by:

To be general, at a distance r from the centre of mass of a mass M, the gravitational potential is given by:

Gravitational potential VG and field strength g

The following indicates the force F on a mass m, the gravitational field strength g and the potential VG at r in the gravitational field due to mass M:

Consider the above diagram, representing mass m in the gravitational field of mass M. The field around M is radial, i.e. it is the same in all directions. We could draw a line from M in any direction representing an axis, and the variation of the field along all such axes would be the same. In the diagram, the axis has been taken as being to the right. It is conventional to take the direction away from a field source as being positive. This means that F and g in the diagram are negative, since they act to the left, towards M. Now, the slope of a graph going up from left to right, as above, is positive. Hence, to indicate the direction of g, we say that:

(return to start of page)

(return to CONTENTS)

© A


A Level Physics - Copyright © A C Haynes 1999 & 2004