Question

Compared to the strength of Earth's gravity at its surface r = Rę where Re is the radius of the Earth, how much weaker is gravity at a distance of r = 12RE?

Answer 1

Gravitational force on an object at the given distance would be
`0.007` times the value of the force on the same object at the surface.

Step-by-step explanation:

For a point mass above the surface of the Earth (a sphere), magnitude the gravitational attraction on the point mass can be found with the equation:

`\displaystyle F = (G\, M\, m)/(r^(2))`,

Where:

• 
  `G` is the gravitational constant,
• 
  `M` is the mass of the Earth,
• 
  `m` is the mass of the object above the surface of the Earth, and
• 
  `r` is the distance between the object and the center of mass of the Earth.

In this equation, observe that if all other values stay the same, magnitude of gravitational attraction would be proportional to
`(1/r^(2))`. In other words, magnitude of the attraction would be inversely proportional to the square of the distance between the object and the center of mass of the Earth.

In this question, it is given that the distance between the object and the center of the Earth is now
`12` times the value at the surface. The gravitational attraction would be
`(1/12^(2)) \approx 0.007` times the value at the surface.