Set 7 Problem number 4

The acceleration of gravity at the surface of the Earth is 9.8 m/s^2 and the the radius of the Earth is approximately 6400 km.

Using proportionality, find the field strength at twice the radius of the Earth from its center and the field strength at 4 times the radius of the Earth from its center.

Find the strength at a distance of 24700 kilometers from its center.

Find the distance from Earth's center where the gravitational field of the Earth is half its value at the surface.

Solution

At double the Earth's radius from its center, the area of the sphere over which gravity is spread is 4 times as great as at the surface, since the area of a sphere is proportional to the square of its radius.

The strength of the gravitational field a twice the Earth's radius will therefore be 1/4 as great, or 1/4 * (9.8 m/s ^ 2) = 2.45 m/s ^ 2.
At four times the radius of the Earth, the area will be 16 times as great, so the gravitational field strength will be 1/16(9.8 m/s ^ 2) = .6125 m/s^2.

At 24700 kilometers from the center of the Earth, the ratio of the radii of the spheres is

ratio of radii = (6400 / 24700 ),

so the field strength at this radius will be

field strength at 24700 km = 9.8 m/s ^ 2 * (6400 / 24700 ) ^ 2 = .65 m/s ^ 2.

To find the radius at which the gravitational field falls to half its value at the surface, we let R stand for this radius.

The ratio of radii will then be (6400 km / R), and the field strength will be 9.8 m/s ^ 2/(6400 km / R) ^ 2.
The field will fall to half when its strength is (1/2) * (9.8 m/s ^ 2) = 4.9 m/s ^ 2, so we have

9.8 m/s ^ 2 * (6400 km / R) ^ 2 = 4.9 m/s ^ 2.

Solving this equation for R, we obtain R = 9050 kilometers, which is about 1.4 Earth radii from the center.

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.