Question

The radius in kilometers, to which the present radius of the earth ($R{\text{ }} = {\text{ }}6400km$) is to be compressed so that the escape velocity is increased ten times is:
A. 6.4
B. 64
C. 640
D. 4800

Answer 1

Hint: The escape velocity of earth is the velocity required by the object to escape the gravitational field of earth. It is inversely proportional to the square root of the radius of Earth. Using this relation, we can find the radius to which the Earth is to be compressed, such that the radius of Earth is ten times the initial escape velocity of Earth.

Formula used:
$\dfrac{{{V_{e1}}}}{{{V_{e2}}}} = \sqrt {\dfrac{{{R_{e2}}}}{{{R_{e1}}}}} $

Complete answer:
Escape velocity is the velocity required to escape the Earth’s gravity field. It is mathematically given by
${V_e} = \sqrt {\dfrac{{2GM}}{R}} $

Where
${V_e}$ is the escape velocity of Earth
$G$ is the gravitational constant
$M$ is the mass of the Earth
$R$ is the radius of Earth

From the above relation we can understand that
${V_e} \propto \dfrac{1}{{\sqrt R }}$
Using this we, deduce the relation
$\dfrac{{{V_{e1}}}}{{{V_{e2}}}} = \sqrt {\dfrac{{{R_{e2}}}}{{{R_{e1}}}}} $
Here,
${V_{e1}}$ is the initial escape velocity of Earth
${R_{e1}}$ is the initial radius of Earth
${V_{e2}}$ is the final escape velocity of Earth
${R_{e2}}$ is the final radius of Earth

It is mentioned in the question, that ${V_{e2}} = 10\left( {{V_{e1}}} \right)$.

Let, ${R_{e2}} = x\left( {{R_{e1}}} \right)$ where x is the no. of times the initial radius is compressed to. Substituting these in the formula
We have,
$\eqalign{
  & \dfrac{{{V_{e1}}}}{{{V_{e2}}}} = \sqrt {\dfrac{{{R_{e2}}}}{{{R_{e1}}}}} \cr
  & \Rightarrow \dfrac{{{V_{e1}}}}{{10({V_{e1}})}} = \sqrt {\dfrac{{x\left( {{R_{e1}}} \right)}}{{{R_{e1}}}}} \cr
  & \Rightarrow x = \dfrac{1}{{100}} \cr} $

Now, we have the new radius as
$\eqalign{
  & {R_{e2}} = x\left( {{R_{e1}}} \right) = \dfrac{{6400km}}{{100}} = 64km \cr
  & \therefore {R_{e2}} = 64km \cr} $

So, the correct answer is “Option B”.

Note:
In the question, nothing is mentioned about mass. For this reason, we assume that the mass is constant throughout the problem. The student should look out if change in mass is mentioned, as it might affect the solution. In that case, the formula is as follows
$\dfrac{{{V_{e1}}}}{{{V_{e2}}}} = \sqrt {\dfrac{{{R_{e2}}}}{{{R_{e1}}}} \times \dfrac{{{M_{e1}}}}{{{M_{e2}}}}} $