Question

what will be the value of g at height 3200 km above the Earth's surface​

Answer 1

Answer:

Assume-

     h= 3200km = 3200000m = 3.2 x10^6

     R= 6400km = 6400000m = 6.4 x 10^6

     g= 9.8m/s^2

We know that-

           g'= g[R^2/(R+h)^2]

              = 9.8(40.96 x 10^12/92.16 x 10^12)

              = 9.8 x 4/9

              = 4.355m/s^2 ~ 4.4m/s^2

Thus, value of g at a place 3200km above surface of the earth is 4.4m/s^2.

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Answer 2

Answer:

Given:

Height = 3200 km

To find:

Value of acceleration due to gravity

Formulas used:

Let new gravity be g"

$g" \: = \frac{g}{ {(1 + \frac{h}{r} )}^{2} }$

g => gravity on Earth surface

h => height

r => radius of Earth = 6400 km

Calculation:

Putting the values as given in the question :

$g" \: = \frac{g}{ {(1 + \frac{3200}{6400} )}^{2} }$

$g" \: = \frac{g}{ {(1 + \frac{1}{2} )}^{2} }$

$g" \: = \frac{g}{ {( \frac{3}{2} )}^{2} }$

$g" \: = \frac{4g}{9}$

So the final answer is

$\boxed{g" \: = \frac{4g}{9}}$

If we put g = 10 m/s² , then we get :

g" = 4.44 m/s²