Question

The escape velocity on the planet is v. If the radius of the planet contracts to ${\dfrac{{\text{1}}}{{\text{4}}}^{{\text{th}}}}$of present value without any change in its mass, the escape velocity will be
A. Halved
B. Doubled
C. Quadrupled
D. Becomes one fourth

Answer 1

Escape velocity is the minimum velocity required by a body to be projected to overcome the gravitational pull of the earth. The formula for escape velocity is ${\text{v = }}\sqrt {\dfrac{{{\text{2GM}}}}{{\text{R}}}}$. Now substituting the value of R as per question in the formula and finally we can get the value of new escape velocity. As v is inversely proportional to the R. The value of R decreases then eventually the value of v will increase.

Complete step by step solution:
The formula for escape velocity on a planet is:
${\text{v = }}\sqrt {\dfrac{{{\text{2GM}}}}{{\text{R}}}}$
Where v = escape velocity
G = universal gravitational constant whose value is
R = Radius of the planet or the distance from the centre of the mass of the planet
M = mass of the body to be escaped from or the mass of the planet

According to the question, the radius of the planet contracts to ${\dfrac{{\text{1}}}{{\text{4}}}^{{\text{th}}}}$of present value without any change in its mass.
$\therefore {\text{ R' = }}\dfrac{{\text{R}}}{{\text{4}}}$
And ${\text{M' = M}}$
Now, new formula for escape velocity is given as:
${\text{v' = }}\sqrt {\dfrac{{{\text{2GM'}}}}{{{\text{R'}}}}}$

Now substituting the value of ${\text{M'}}$ and ${\text{R'}}$in the above formula for escape velocity, we get

$${\text{v' = }}\sqrt {\dfrac{{{\text{2GM}}}}{{\dfrac{{\text{R}}}{{\text{4}}}}}} \\\ \Rightarrow {\text{ v' = }}\sqrt {\dfrac{{{\text{4}} \times {\text{2GM}}}}{R}} \\\ {\text{But }}\sqrt {\dfrac{{{\text{2GM}}}}{{\text{R}}}} {\text{ = v}} \\\$$

$\therefore {\text{ v' = 2v}}$
So, the escape velocity will be doubled.

Therefore, option (B) is the correct choice.

Note:
Escape velocity is actually a speed not a velocity because escaped velocity does not specify direction that means no matter what the direction of the object or body in which it travels, the object or body can escape the gravitational field (provided its path does not intersect the planet). The escape velocity from Earth's surface is about ${\text{11,186 m/s}}$