Question

An object has an altitude of 2 times the earth's radius, re and experiences some force of gravity, f. if the object's altitude is doubled, then the new force of gravity will be

Answer 1

When an object's altitude above Earth is doubled from 2 times to 4 times Earth's radius, the new force of gravity experienced by the object will be one fourth of the original force, according to the inverse square law of gravitation.

Step-by-step explanation:

The question involves understanding gravitational force and how it changes with the altitude of an object above a planet. Given an object at an altitude of 2 times Earth's radius (re), experiencing a gravitational force f, when we double this altitude, the new force of gravity can be determined by the inverse square law of gravitation. The law states that the gravitational force is inversely proportional to the square of the distance from the center of the mass causing the gravitational attraction. Therefore, if the altitude is doubled, meaning the object is now at a distance of 4 re from the center of the Earth, the force of gravity would be (1/2)2 or 1/4 of the initial force f.

Answer 2

Note: I assume the altitude is measured from Earth's center (it's not clear from the text if it's from the Earth's center or from the surface)

The force of gravity experienced by the object is given by

`F=G (Mm)/(r^2)`
where
G is the gravitational constant
M is the mass of the Earth
m is the mass of the object
r is the distance of the object from Earth's center

Initially, its distance from Earth's center is twice the Earth radius:
`r=2 r_e`, so the force of gravity is

`F=G (Mm)/((2r_e)^2)=G (Mm)/(4 r_e^2)`

Later, the object altitude is doubled, so its new distance from Earth's center is 4 times the Earth's radius:

`r'=4 r_e`
and so the new force of gravity is

`F'=G (Mm)/((4 r_e)^2)=G (Mm)/(16 r_e^2)= (1)/(4) G (Mm)/(4 r_e^2) = (1)/(4)F`
therefore, the new force is 1/4 of the original force.