A body weighs 72 N on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth?

Question

Question

This question was previously asked in

NEET 2020 Official Paper (Held On: 13 September, 2020)

Answer 1

Concept:

As the height of body from surface of earth increases , value of acceleration due to gravity decreases.

Value of g on the surface of earth, g = 9.8 m s^-2.

Value of g at a height h above the surface of earth is \(g' = {g \over(1+ {h\over R})^2}\)

Where R is radius of earth.

Calculation:

\(m{g_h} = \frac{{m{g_0}}}{{{{\left( {1 + \frac{h}{R}} \right)}^2}}}\)

\(W = \frac{{72}}{{{{\left( {1 + \frac{{R/2}}{R}} \right)}^2}}}\)

\(W = \frac{{72}}{{{{(3/2)}^2}}} = \frac{4}{9} \times 72 = 32~N\)