Question

A rocket is launched vertically from the surface of earth with an initial velocity $\nu$. How far above the surface of earth it will go? Neglect the air resistance.

• A. $R{(\frac{2gR}{{\nu }^2}-1)}^{-1/2}$
• B. $R(\frac{2gR}{{\nu }^2}-1)$
• C. $R{(\frac{2gR}{{\nu }^2}-1)}^{-1}$
• D. $R{(\frac{2gR}{{\nu }^2}-1)}^2$

Answer 1

Answer: (C)

$R{(\frac{2gR}{{\nu }^2}-1)}^{-1}$

Let $r$ be the distance from center of earth the rocket goes.

From conservation of energy:

$GMm(\frac{1}{R}-\frac{1}{r})=\frac{1}{2}m{v}^2$

$⟹\frac{1}{r}=\frac{1}{R}-\frac{v^2}{2GM}$

$⟹r=\frac{2GMR}{2GM-R{v}^2}$

Therefore the height upto which the rocket goes:

$=r-R=R\left(\frac{R{v}^2}{2GM-R{v}^2}\right)$

$=R(\frac{2GM}{R{v}^2}-1{)}^{-1}$

$=R(\frac{2gR}{v^2}-1{)}^{-1}$