Problem: The gravitational effect of the Earth can be thought of as being spread out over larger and larger spheres, each concentric with the Earth. The greater the area over which the field is spread, the less the strength of the field.
At the radius of the Earth, the field strength is measured to be 9.8 m/s ^ 2. What would be the field strength at twice the radius of the Earth?
What would the strength be at 4 times the radius of the Earth?
If the radius of the Earth is 6400 km, then what would be the strength at a distance of 2.570 * 10 ^ 4 kilometers from its center?
At what distance from its center does the gravitational field of the Earth fall to half its value at the surface?
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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 will therefore be 1/4 as great, or ¬ (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 2.570 * 10 ^ 4 kilometers from the center of the Earth, the ratio of the radii of the spheres is (6400 / 2.570 * 10 ^ 4), so the field strength at this radius will be 9.8 m/s ^ 2 * (6400 / 2.570 * 10 ^ 4) ^ 2 = .6077 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. We thus 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.
Generalized Response: 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.
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Figure Description: 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 turns out to be 4 `pi G M, where G = 6.67 * 10^-11 N m^2 / kg^2. When divided by the area of the sphere concentric with the center of mass of mass M, we obtain the gravitational field strength, or gravitational acceleration, at the surface of that sphere.
At distance r from the center of Earth, then, the area ratio of the two spheres would be (r / Re)^2, where Re stands for Earth radius. The gravitational field strengths would then have ratio (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.
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