Question

A rocket is launched vertically upward from the surface of the earth with an initial velocity of $ 10 \,km/s $ . If the radius of the earth is $ 6400 \,km $ 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 $= 6400\, km$
According to the question,
$- \frac{GMm}{R} + \frac{1}{2} mv^{2} = \frac{GMm}{\left(R + h\right)}$
$ \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 \times10^{4} \,km$