Question

At what distance above the surface of the Earth is the acceleration due to earth's gravity .980m/s^2 if the acceleration due to gravity at the surface has magnitude 9.80 m/s^2? Draw a free-body diagram.

Answer 1

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Answer 2

SOLUTION :

Let that height be h  m from the earth’s surface at which altitude, g’ = 0.980 m/s^2

g at the earth’s surface is = 9.8 m/s^2.

As we go up, value of g decreases.

Value of g’ at height h is given by the following formula:

g’ = g (1 - 2h / R) 

Where,

g’ = acceleration due to gravity at height h = 0.980  m/s^2 (given in the problem)

g = acceleration due to gravity at earth’s surface = 9.8 m/s^2

h = height at which g’ = 0.980 m/s^2. (To be determined)

R = radius of the earth = 6371 km = 6.371*10^6 m

So, as per above formula :

0.980 = 9.8( 1 - 2 h / (6.371 * 10^6) )

=> h = (1 - 0.980/9.8) *  (6.371 * 10^6) / 2 

=> h = 2 866 950 m = 2866. 95 km

=> h = height at which g’ is 0.980 m/s^2 = 2866.95 km   (ANSWER).

Free body diagram :

            0      object at height 2866.95 km from earth's surface

            I.  I 

            I   V    g' = 0.980 m/s^2

            I  

         ____ Earth's surface