Question

A particle is fired vertically upward with a speed of 9.8 km/s. Find the maximum height attained by the particle. Radius of earth = 6400 km and g at the surface = 9.8m/s${}^2$. Consider only earth's gravitation.

• A. 20000 km.
• B. 20900 km.
• C. 1900 km.
• D. 209 km.

Answer 1

The correct option is B

20900 km.

At the surface of the earth, the potential energy of the earth-particle system is $-\frac{GMm}{R}$ with usual symbols.

The kinetic energy is $\frac{1}{2}m{v}_0^2$ where $v_0=9.8$ km/s. At the maximum height the kinetic energy is zero.

If the maximum height reached is H, the potential energy of the earth-particle system at this instant is $\frac{GMm}{R+H}$. Using conservation of energy,

$-\frac{GMm}{R}+\frac{1}{2}m{v}_0^2=-\frac{GMm}{R+H}.$

Writing GM=gR${}^2$ and dividing by m,

$-gR+\frac{v_0^2}{2}=\frac{-g{R}^2}{R+H}$

Or,$\frac{R^2}{R+H}=R-\frac{v_0^2}{2g}$

Or,$R+H=\frac{R^2}{R-\frac{v_0^2}{2g}}$

Putting the values of R, $v_0$ and g on the right side,

$R+H=\frac{(6400km{)}^2}{6400km-\frac{(9.8km/s{)}^2}{2\times 9.8m/s^2}}$

$=\frac{(6400km{)}^2}{1500km}=27300km$

Or,$H=(27300-6400)km=20900km.$