Question

The escape velocity of the earth is $11.2\frac{km}{s}$. What is the escape velocity on a planet whose radius is twice that of the earth and mass four times that of the earth?

Answer 1

Step 1: Given data

Escape Velocity ${\left(v_e\right)}_1=11.2\frac{km}{s}$

Step 2: Calculating the escape velocity on a planet whose radius is twice that of the earth and mass four times that of the earth.

Escape velocity is the lowest velocity that a body must have in order to escape the gravitational attraction of a particular planet or another object.

${\left(v_e\right)}_1=\sqrt{\frac{2GM}{r}}......................\left(i\right)$

According to the given condition, escape velocity is,

${\left(v_e\right)}_2=\sqrt{\frac{2G\left(4M\right)}{2r}}$

${\left(v_e\right)}_2=\sqrt{2}\sqrt{\frac{2GM}{r}}...........\left(ii\right)$

Step 3: By substituting the values from equation (i) in equation (ii), we get

${\left(v_e\right)}_2=\sqrt{2}{\left(v_e\right)}_1$

${\left(v_e\right)}_2=\sqrt{2}\times 11.2$

Step 4: Final answer

${\left(v_e\right)}_2=15.83\frac{km}{s}$

Hence the escape velocity will be $15.83\frac{km}{s}$