The escape velocity of a body on the surface of the earth is $11.2\, km/s^{-1}$. If the earth's mass increases to twice its present value and radius of the earth becomes half, the escape velocity becomes

Question

Two identical particles each of mass $ m $ go around a circle of radius $ a $ under the action of their mutual gravitational attraction. The angular speed of each particle will be:

Answer 1

The escape velocity from a celestial body is given by the formula:

$v_e = \sqrt{\dfrac{2GM}{R}}$

where

$v_e$ is the escape velocity,
$G$ is the gravitational constant,
$M$ is the mass of the celestial body, and
$R$ is the radius of the celestial body.

Given that the escape velocity from the Earth's surface is $11.2\, \text{km/s}$, we are to find the new escape velocity if the Earth's mass becomes twice its present value ($2M$) and its radius becomes half ($\dfrac{R}{2}$).

Substituting these new values in the escape velocity formula, we have:

$v'_e = \sqrt{\dfrac{2G(2M)}{\dfrac{R}{2}}} = \sqrt{\dfrac{4GM}{\dfrac{R}{2}}} = \sqrt{\dfrac{8GM}{R}}$

Since the original escape velocity is:

$v_e = \sqrt{\dfrac{2GM}{R}} = 11.2\, \text{km/s}$

Comparing the two expressions, we find:

$v'_e = \sqrt{4} \times v_e = 2 \times 11.2 = 22.4\, \text{km/s}$

Therefore, if the Earth's mass were to double and its radius halved, the escape velocity would become $22.4\, \text{km/s}$. Thus, the correct answer is $22.4\, \text{km/s}$.

The escape velocity of a body on the surface of the earth is $11.2\, km/s^{-1}$. If the earth's mass increases to twice its present value and radius of the earth becomes half, the escape velocity becomes

The Correct Option is A

Solution and Explanation

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