Question

A rocket is launched vertically upward from the surface of the earth with an initial velocity of $10km/s$ . If the radius of the earth is $6400km$ and atmospheric resistance is negligible. Find the distance above the surface of the earth that the rocket will go.

• A. $2.5\times 10^{4}km$
• B. $3.0\times 10^{4}km$
• C. $4.0\times 10^{3}km$
• D. $3.0\times 10^{3}km$

Answer 1

Answer: (A)

Radius of earth $=6400km$
According to the question,
$-\frac{GMm}{R}+\frac{1}{2}m{v}^2=\frac{GMm}{\left(R+h\right)}$
$\frac{-GM}{R}+\frac{1}{2}{v}^2=\frac{-GM}{R+h}$
$\frac{1}{2}{v}^2=\frac{-GM}{\left(R+h\right)}+\frac{GM}{R}$
$\frac{1}{2}{v}^2=GM\left(\frac{1}{R+h}-\frac{1}{R}\right)$
$\frac{1}{2}\times {\left(10\right)}^2=6.6\times 10^{-11}\times 6.0\times 10^{24}$
$\left[\frac{1}{6400+h}-\frac{1}{6400}\right]$
$h=2.5\times 10^{4}km$