Question

At what height the acceleration due to gravity is 25% of that at the surface of earth, in terms of radius of the earth ?

Answer 1

Answer:

To find :

Height (h) at which acceleration due to gravity(g) will be 25% of that at the surface of the Earth.

Solution :

Let the acceleration due to gravity at the surface of the earth be g₀.

We know that, acceleration due to gravity at the surface of the earth is given by :

        g₀ = $= \frac{GM}{R^{2} }$

       

      where G is the universal gravitational constant, M is the mass of the Earth and R is the radius of the Earth.

Acceleration due to gravity at a height h above the earth's surface is given by :

       

As per the given condition, g is 25% of g₀.

∴ g = 0.25 ×g₀.

$\frac{gm}{(r + h {)}^{2} } = \frac{1}{4} \frac{(gm)}{ {(r)}^{2} }$

$∴ \frac{R^{2} }{(R+h)^{2} } = \frac{1}{4}$

$∴ \frac{R}{R+h} = \frac{1}{2}$

$∴ 2R = R + h$

$∴ R = h$

Answer : The acceleration due to Gravity will be 25% of that at the surface of the Earth at height of R.