Set 7 Problem number 6

The strength of the gravitational attraction between any two masses m1 and m2, with their centers of mass separated by distance r, is the force

gravitational attraction F = G m1 m2 / r ^ 2,

where G is a proportionality constant approximately equal to 6.67 * 10^-11 N m ^ 2/kg ^ 2.

Give the strength of the gravitational attraction felt by a human being of mass 84 kg to a rock sphere with radius 2.3 km and density 2.7 times that of water (water's density is 1000 kg/m ^ 3), assuming that the person's entire mass is right at the surface of the sphere.

Give the attraction if the sphere was compressed to a radius of 230 meters, and if it was compressed to radius 23 meters.
To what radius would the sphere have to be compressed in order to exert a force equal to the weight of this individual on the surface of the Earth?

Solution

In order to find the force F = G m1 m2 / r^2, we need to know both masses m1 and m2 and the distance r between the centers of mass.

We know the mass of the person (call it m2), and if the person is hugging the sphere, the distance r between the centers of mass will be close to the radius of the sphere.

We need to find the mass m1 of the sphere.

We know that the density is 2.7 * 1000 kg/m^3.

If we knew the volume of the sphere, we could then find its mass.

The radius of the sphere is 2.3 kilometers = 2300 meters.

Its volume is therefore (4 `pi /3)r^3 = 5.096513E+10 cubic meters.

Multiplying this volume by the density 2700 kg/m^3 we obtain the sphere's total mass

sphere mass m1 = 1.376059E+14 kilograms.

We can now find the gravitational force.

We substitute m1 = 84 kg, m2 = 1.376059E+14 kilograms r = 2300 meters into F = G m1 m2 / r ^ 2, and use the universal gravitational constant G = 6.67 * 10^-11 N m ^ 2/kg ^ 2 to obtain force

F = .145 Newtons.

If the sphere is compressed to a radius of 230 meters, its mass remains the same, as does the mass of the person.

So we use the same masses m1 and m2, the same value of G (since it is a universal constant), and the new value 230 meters for r to obtain

force F = 14.574 Newtons.

Note that this is 100 times the force experienced at the first radius. The person is 10 times closer to the mass, so r ^ 2 is 1/10 as great, so the gravitational influence of the mass is spread over the surface of an imaginary sphere with 1/10 the radius.

This imaginary sphere has 1/100 times the area of the first, so the gravitational field will be 100 times as intense. The force on the person will therefore be 100 times as great.

When the original sphere is compressed to radius 23 meters, which is 1/100 its original radius, a proportionality argument similar to the above tells us that the gravitational field will become 100^2 = 10,000 times as intense.

This is confirmed by substituting the original masses, with the new radius r = 23 meters, into F = G m1 m2 / r ^ 2.

We obtain F = 1457.425 Newtons, which is in fact 10000 times as great as the force found at the original radius.

The weight of the individual on Earth is 9.8 m/s ^ 2 ( 84 kg) = 823.2 Newtons.

In order to find the radius at which F = G m1 m2 / r ^ 2 is equal to this weight, we solve this equation for r to obtain r = `sqrt[G m1 m2 / F].

Substituting the known values of m1 and m2 and the desired value of F, we obtain r = 30.6033 meters.

If all the mass in a sphere of radius R is compressed into a sphere of radius r, then the gravitational influence of the mass will be concentrated in a smaller area and the gravitational field will therefore be greater. The ratio of areas will be the square of the ratio of radii:

area ratio = (r / R) ^ 2.

The ratio of gravitational fields will therefore be the inversse of the area ratio:

ratio of gravitational fields = (R / r) ^ 2.