The escape velocity of a body on the surface of the earth is 11.2km /s . If the earth’s mass increases to twice its present value and the radius of the earth becomes half, the escape velocity would become

The correct option is 22.4 km / s.

To understand how the escape velocity changes, let us first recall the formula for the escape velocity ($v_e$) of a body from the surface of a celestial body (like Earth):
$v_e=\sqrt{\frac{2GM}{R}}$
where:
- $G$ is the universal gravitational constant,
- $M$ is the mass of the Earth, and
- $R$ is the radius of the Earth.

According to the problem, the initial escape velocity is:
$v_e=11.2\text{km/s}$

Now, let the new mass of the Earth be $M^{\prime }$ and the new radius be $R^{\prime }$. According to the given conditions:
- The mass increases to twice its present value: $M^{\prime }=2M$
- The radius becomes half of its present value: $R^{\prime }=\frac{R}{2}$

Let us write the expression for the new escape velocity ($v_e^{\prime }$):
$v_e^{\prime }=\sqrt{\frac{2G{M}^{\prime }}{R^{\prime }}}$

Substitute the values of $M^{\prime }$ and $R^{\prime }$ into the new equation:
$v_e^{\prime }=\sqrt{\frac{2G\left(2M\right)}{\left(\frac{R}{2}\right)}}$
$v_e^{\prime }=\sqrt{\frac{4·2GM}{R}}$
$v_e^{\prime }=2\sqrt{\frac{2GM}{R}}$

Since $\sqrt{\frac{2GM}{R}}=v_e$, we can write:
$v_e^{\prime }=2v_e$

Now, substituting the initial value of the escape velocity ($v_e=11.2\text{km/s}$):
$v_e^{\prime }=2\times 11.2\text{km/s}=22.4\text{km/s}$

Thus, the new escape velocity would become 22.4 km / s.