Set 7 Problem number 5

Use the inverse square proportionality g = k/r ^ 2 and the field strength at the surface of the Earth to obtain an expression for the gravitational field strength of the Earth at distance r from its center, assuming that r is at least as great as the 6400 km radius of the Earth.

Give the proportionality constant k, then use your result to calculate the field strength at distances 20000, 200000 and 2000000 kilometers from the center of the Earth.

Solution

We assume the proportionality g = k/r ^ 2.

Substituting the known field strength g = 9.8 m/s ^ 2 at distance 6400 kilometers, we obtain 9.8 m/s ^ 2 = k / (6400 km) ^ 2.
Solving for k we obtain k = 4.01408E+08 km m / s ^ 2.
This gives us the equation g = [ 4.01408E+08 km m / s ^ 2] / r ^ 2.

We substitute the given radii into the equation g = [ 4.01408E+08 km m / s ^ 2] / r ^ 2.

Substituting r = 20000 kilometers we obtain

field strength at 20000 km: g = 4.01408E+08 kg m /s ^ 2 / ( 20000 km) ^ 2 = 1.003 m/s ^ 2.

We substitute the other two distances and obtain

field strengths at 200000 km and 2000000 km: .01003 m/s ^ 2 and .0001003 m/s ^ 2.

[Note that we could have found these results by using the fact that the second radius is 10 times that of the first and the third is 100 times that of the first. A little reflection tells us that the spheres over which the gravitational effect is spread will have areas which are respectively 100 and 10000 times as great as that of the first. The fields will therefore have 1/100 and 1/10000 times the strength of the first. These results coincide with the results of substitution.]

Generalized Solution

If a planet has radius R and gravitational field strength g at its surface, then the inverse square proportionality dictates that at radius r1 the field strength will be (R / r1) ^ 2 times as great as at the surface, or

field strength at distance r1 from center = (R / r1) ^ 2 * g.

The distance at which the field strength falls to half its value at the surface is therefore the distance r1 at which we have

(R / r1) ^ 2 * g = (1/2) * g.

We can easily solve this for r1, obtaining

r1 = `sqrt(2) * R, or approximately 1.414 R.

Explanation in terms of Figure(s), Extension

The figure below shows a black dot representing Earth, a blue sphere of radius r1 and a green sphere of radius r2.

The gravitational effect of the Earth spreads uniformly, with the same total effect spread out over each sphere.
The effect is therefore spread more thinly over the largest sphere then over the smallest.

The gravitational acceleration at the surface of any sphere is equal to the 'area density' of the total gravitational effect.

This total effect is known to be 4 `pi G M, where G = 6.67 * 10^-11 N m^2 / kg^2 and M is the mass of the planet.
When divided by the area of any sphere concentric with the planet we obtain the gravitational field strength, or gravitational acceleration, at the surface of that sphere.
If the sphere has radius R, the area density is 4 `pi G M / (4 `pi R^2) = G M / R^2.

At distance r from the center of Earth, the area ratio of the two spheres would be (r / Re)^2, where Re stands for Earth radius.

The ratio of the gravitational field strength at R and at the surface of the Earth would thus be (Re / R) ^ 2, so that the strength at distance R would be (Re / R) ^ 2 * g, where g is the strength at the surface of the Earth.
At radius R = r1 the strength would be (Re / r1) ^ 2 * g; at R = r2 the strength would be (Re / r2) ^ 2 * g.