Question

A satellite is one Earth radius above the surface of Earth. How does teh acceleration due to gravity at that location compare to acceleration at the surface of Earth?

Answer 1

Answer:

Explanation:

Given

Satellite is at distance equal to earth radius from earth surface

$r_{eff}=r+r=2 r$

Where r=radius of earth

acceleration due to gravity is given by

$g=\frac{GM}{r^2}$

for $r=2 r$

$g'=\frac{GM}{(2r)^2}$

$g'=\frac{GM}{r^2}\times 0.25$

$g'=0.25 g$

Answer 2

The acceleration due to gravity will be "0.25 g".

Acceleration and Gravity:

Whenever an item falls spontaneously from either a particular height to the ground atmosphere, its velocity fluctuates, and this change in velocity causes acceleration throughout the entity or object, which is known called acceleration due to gravity.

Let,

The radius of earth "r".

Now,

The radius from earth surface be:

$r_{eff}$ = r + r

      = 2r

As we know,

The acceleration due to gravity be:

→ g = $\frac{GM}{r^2}$

For, r = 2r

  g' = $\frac{GM}{(2r)^2}$

      = $\frac{GM}{r^2}\times 0.25$

      = 0.25 g

Thus the above answer is correct.        

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