Question

The escape velocity on earth is 11.2 km/s. If the mass and radius of the Earth are doubled, then the new value of its orbital velocity is _________.

• A. $\frac{11.2}{\sqrt{2}}km/s$
• B. $\left(\sqrt{2}\right)\left(11.2\right)km/s$
• C. $5.6\left(\sqrt{2}\right)km/s$
• D. $\frac{5.6}{\sqrt{2}}km/s$

Answer 1

The correct option is A

$\frac{11.2}{\sqrt{2}}km/s$

Step 1, Given data

Escape velocity on earth $V_e=11.2\raisebox{1ex}{$km$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

Mass and radius of the earth become double

Step 2, Finding the new orbital velocity

If the escape velocity is 11.2 km/s the orbital velocity $V_0=\frac{V_e}{\sqrt{2}}=\frac{11.2}{\sqrt{2}}\raisebox{1ex}{$km$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Escape velocity is given as $V_e=\sqrt{\frac{2GM}{R}}$

Where, G = universal constant, M = Mass of the earth, R = Radius of the earth

Mass of the earth and radius become half

So, $V_e=\sqrt{\frac{2GM}{R}},V_e\text{'}=\sqrt{\frac{2G{\displaystyle 2M}}{{\displaystyle 2R}}}=\sqrt{\frac{2GM}{R}}=11.2km/s$

Now, New orbital velocity $V_o\text{'}=\frac{V_e\text{'}}{\sqrt{2}}=\frac{11.2}{\sqrt{2}}km/s$

Therefore the new orbital velocity is $\frac{11.2}{\sqrt{2}}km/s$

Hence the correct option is (A) $\frac{11.2}{\sqrt{2}}km/s$