Question

The kinetic energy needed to project a body of mass $\mathrm{m}$ from the earth's surface to infinity is?

• A. $\mathrm{mgR}/2$
• B. $\mathrm{mgR}/4$
• C. $\mathrm{mgR}$
• D. $2\mathrm{mgR}$

Answer 1

The correct option is C

$\mathrm{mgR}$

Explanation for the correct option:

Option(C):$\mathrm{mgR}$

Step 1: Given data:

The mass of the body is given as= $\mathrm{m}$

Step 2: Formula for calculating kinetic energy:

If any object through with escaping velocity, it will never come back to the ground, So we can assume it will reach an infinite distance.

Hence for this, the kinetic energy of an object which is a measure of the work an object can do by virtue of its motion can be given as:

$\mathrm{K}.\mathrm{E}=\frac{1}{2}\mathrm{m}{{\mathrm{V}}_{\mathrm{e}}}^2$

Where, ${\mathrm{V}}_{\mathrm{e}}$ is termed as the escaping velocity, which can be given as:

${\mathrm{V}}_{\mathrm{e}}=\sqrt{2\mathrm{gR}}$

(Where $\mathrm{R}$ is the radius of the earth)

Step 3: Calculation of kinetic energy:

Hence, by putting the value of ${\mathrm{V}}_{\mathrm{e}}$, in the given equation, we can calculate the value of kinetic energy as:

$$\begin{gathered} \mathrm{K}.\mathrm{E}=\frac{1}{2}\mathrm{m}{{\mathrm{V}}_{\mathrm{e}}}^2 \\ \Rightarrow \mathrm{K}.\mathrm{E}=\frac{1}{2}\mathrm{m}\left(2\mathrm{gR}\right) \\ \Rightarrow \mathrm{K}.\mathrm{E}=\mathrm{mgR} \end{gathered}$$

Hence, the correct option is C.