Question

An Earth satellite is orbiting at a distance from the Earth’s surface equal to one Earth radius (4 000 miles). At this location, the acceleration due to gravity is what factor times the value of gat the Earth’s surface?
A. There is no acceleration since the satellite is in orbit.
B. 2
C. 1/2
D. 1/4

Answer 1

Answer:

The acceleration due to gravity is $\dfrac{1}{4}$ times the value of g at the Earth’s surface.

(D) is correct option.

Explanation:

Given that,

Radius = 4000 miles

We need to calculate the gravitational force at surface

Gravitational force on the mass m on the surface of the earth

At r = R

$F=mg$

$mg=\dfrac{GmM}{R^2}$....(I)

We need to calculate the gravitational force at height

Gravitational force on a mass m from the center of the earth,

At r = R + R = 2 R

$F'=mg'$

$mg'=\dfrac{GmM}{4R^2}$....(II)

Dividing equation (II) by equation (I)

$\dfrac{mg'}{mg}=\dfrac{\dfrac{GmM}{4R^2}}{\dfrac{GmM}{R^2}}$

$\dfrac{g'}{g}=\dfrac{1}{4}$

Hence, The acceleration due to gravity is $\dfrac{1}{4}$ times the value of g at the Earth’s surface.