Question

An object is projected vertically upward from the surface of the earth of mass M with a velocity such that the maximum height reached is eight times the radius R of the earth. The speed at half the maximum height will be

• A. $\frac{2}{3}\sqrt{\frac{2GM}{3R}}$
• B. $\frac{2}{3}\sqrt{\frac{GM}{R}}$
• C. $\frac{2}{3}\sqrt{\frac{2GM}{5R}}$
• D. $\sqrt{\frac{GM}{R}}$

Answer 1

The correct option is C $\frac{2}{3}\sqrt{\frac{2GM}{5R}}$

Using energy balance we have

$\frac{1}{2}m{v}^2=-\frac{GMm}{9R}-(-\frac{GMm}{R})$

Solving we get the relation for velocity of projection as

$v^2=\frac{16GM}{9R}$

Now gravitational potential difference between the point at the surface and half the maximum height is given as

$-\frac{GMm}{5R}-(-\frac{GMm}{R})=\frac{4GMm}{5R}$

This would be equal to the change in Kinetic energy between the same points. Thus we get

$\frac{1}{2}m(v^2-v_h^2)$(here $v_h$ is the velocity at the height of 4R from the earth)

Thus we get

$\frac{1}{2}m(v^2-v_h^2)=\frac{4GMm}{5R}$

or

$v_h^2=\frac{16GM}{9R}-\frac{8GM}{5R}$

or

$v_h=\frac{2}{3}\sqrt{\frac{2GM}{5R}}$