Earth exerts a gravitational force of 432 N on an object on its surface.Earth has radius R.If the object is raised to a height of 3R above Earth's surface,what is the gravitational force that Earth will exert on it?
A)108 N
B)48.0 N
C)27.0 N
D)305 N
E)144 N

Question

Quizplus · Accepted Answer

The gravitational force exerted by the Earth on an object of mass 'm' at the surface of the Earth is given by:
F = G (Mm/R^2)
where G is the universal gravitational constant, M is the mass of the Earth, m is the mass of the object, and R is the radius of the Earth.

Given, F = 432 N, and let R' be the new distance between the center of the Earth and the object (3R).

So, the new gravitational force,
F' = G (Mm/R'^2)

Dividing F' by F,
F'/F = (R/R')^2

F' = F * (R/R')^2

Substituting the given values, we get
F' = 432 * ((1/3)^2)
F' = 432 * (1/9)
F' = 48 N
Therefore, the gravitational force that Earth will exert on the object when it is raised to a height of 3R above Earth's surface is 48 N, which is choice C).

Earth Exerts a Gravitational Force of 432 N on an Object