(a) Find the altitude above the Earth\'s surface where Earth\'s gravitational fi

Question

(a) Find the altitude above the Earth's surface where Earth's gravitational field strength would be three-fourths of its value at the surface. (Assume Re = 6.371 103 km.)

(b) Find the altitude above the Earth's surface where Earth's gravitational field strength would be one-fourth of its value at the surface.
[Hint: First find the radius for each situation; then recall that the altitude is the distance from the surface to a point above the surface. Use proportional reasoning.]

I really need help with this homework problem, please walk me through how you got the answer. Thank you!

Explanation / Answer

The acceleration due to gravity is inversely proportional to the square of the distance from the center of the earth

 

g' = g/r^2

So g' is directly proportional to 1/r^2

(a)  As g becomes 3/4 of the initial value

 

Therefore Distance from the center 2/3 R from the center of the earth.

 

That is the altitude = 2/3 * 6371*10^3 = 7356597.13 m

 

(b)

As the gravitational field strength would be one-fourth of its value i.e,g/4

 

Threfore altitude from the center of the earth = 2R = 12742000 m

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