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)

$2.5 \times 10^4\, km$

Answer 2

Radius of earth $= 6400\, km$
According to the question,
$- \frac{GMm}{R} + \frac{1}{2} mv^{2} = \frac{GMm}{\left(R + h\right)}$
$\frac{-GM}{R} + \frac{1}{2}v^{2} = \frac{-GM}{R+h}$
$\frac{1}{2} v^{2} = \frac{-GM}{\left(R + h\right)} + \frac{GM}{R}$
$\frac{1}{2}v^{2} = GM\left(\frac{1}{R+h} -\frac{1}{R}\right)$
$\frac{1}{2}\times \left(10\right)^{2} = 6.6 \times 10^{-11} \times 6.0 \times 10^{24}$
$\left[\frac{1}{6400+h} - \frac{1}{6400}\right]$
$h = 2.5 \times10^{4} \,km$