Question

A projectile is fired from the surface of earth with initial velocity of $10km/sec$. If the radius of earth is 6400 Km. how high will it go from earth surface

• A. $2500km$
• B. $250km$
• C. $2.5\times {10}^{4}km$
• D. $3.3\times {10}^{4}km$

Answer 1

The correct option is D $3.3\times {10}^{4}km$

The work done for projecting the the body from surface of earth to a height H is
$W={\int }_R^H\frac{GMm}{x^2}dx=GMm[\frac{1}{R}-\frac{1}{H}]$
Here, $\frac{1}{2}m{v}^2=W=GMm[\frac{1}{R}-\frac{1}{H}]$
or $\frac{1}{R}-\frac{1}{H}=\frac{v^2}{2GM}=\frac{{10}^2\times ({10}^3{)}^2}{2\times 6.67\times {10}^{-11}\times 6\times {10}^{24}}=0.12\times {10}^{-5}$
or$\frac{1}{H}=\frac{1}{R}-0.12\times {10}^{-5}=0.015\times {10}^{-5}-0.012\times {10}^{-5}=3\times {10}^{-8}$
$H=3.3\times {10}^{7}m=3.3\times {10}^4$km