# Proof of F = mv2/R

Method A This follows directly from the experiment Sketching a satellite orbit and predicting its period. The mathematics follows directly from the sketch produced in that experiment, reproduced below. This is Newton’s method.

It relies crucially on the crossed chords theorem for a circle, which should be given.

The circle represents the orbit of a satellite of radius R, moving with speed v. The satellite moves from A to B in a time t. Without a force the satellite would have moved to K at constant speed.

Now 'switch-on' gravity and the satellite will fall a distance h in the same time from the tangent from A to the point B. It doesn’t matter whether you let it fall from A first, and then continue in the tangential direction, or vice versa. Anyone who objects that the fall from K to B is not along the radius should look at their scale diagram again: it is almost impossible to see the difference between h and the radial drop. [You may need to talk about 'in the limit'.]

From the crossed chords property, h(2R-h) = x2
but 2R>> h therefore 2h = x2 and so h = x2/2R (equation 1)
now x = AK which is almost the arcAB = vt (equation 2)

Combining 1 and 2, h = (vt)2/2R (equation 3)
h is the vertical fall and so using s = 1/2 at2 = h (equation 4)

Then from (3) and (4) 1/2at2 = (vt)2/2R leading to a = v2/R
Using F = ma then F = mv2/R

The same holds for the motion at all places round the circle. The vertical is always taken to mean the direction from the satellite to the centre of the attracting body.

Method B This method relies on an understanding of vectors.

The circle represents the orbit of a satellite of radius R, moving with speed v
The satellite moves from A to B in a time t

Draw vector AP to represent the initial velocity of the satellite at A, which is along a tangent at A. Draw a second vector of the same length, BQ, to represent the later velocity at B.

Redraw the initial and later vectors, both starting from the same point D. Both have magnitude equal to v. FG, representing the change in velocity, must be added to the old velocity to make the new velocity.

AOB and FDG are similar triangles.

The equation, F = mv2/R, illustrates these relationships:

• the higher the speed v, the bigger the force needed to hold objects in orbit,

so the bigger the central acceleration

• for the same speed, the smaller the radius, or the sharper the curve, the bigger the force and therefore the bigger the acceleration must be.

The acceleration goes up with orbital speed, v, but decreases as the radius, R, increases.

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