Question

The ratio of escape velocity at earth $v_e$ to the escape velocity at a planet $v_p$ whose radius and mean density are twice as that of earth is :

• A. $1 : 2 \sqrt{2}$
• B. $1 : 4$
• C. $1 : \sqrt{2}$
• D. $ 1 : 2 $

Answer 1

The Correct Option is A

 To solve this problem, we need to find the ratio of escape velocities between Earth and a planet with given relative properties: radius and density. The escape velocity from the surface of a celestial body of mass \(M\) and radius \(R\) is given by the formula:

\(v_e = \sqrt{\frac{2GM}{R}}\)

where \(G\) is the universal gravitational constant.

The mass \(M\) of a celestial body can be expressed in terms of its volume and density \(\rho\) as:

\(M = \rho V = \rho \left(\frac{4}{3}\pi R^3\right)\)

Substituting this in the escape velocity formula, we have:

\(v_e = \sqrt{\frac{2G \cdot \rho \cdot \left(\frac{4}{3}\pi R^3\right)}{R}} = \sqrt{\frac{8 \cdot \pi \cdot G \cdot \rho \cdot R^2}{3}}\)

Now let's determine the escape velocities at Earth and the planet:

1. For Earth, let \(\rho_e\) and \(R_e\) be its density and radius. So, the escape velocity \(v_e\\) is:
2. For the planet, with radius \(R_p = 2R_e\) and density \(\rho_p = 2\rho_e\), the escape velocity \(v_p\\) is:

Now, let's find the ratio of escape velocities:

\(\text{Ratio} = \frac{v_e}{v_p} = \frac{v_e}{4v_e} = \frac{1}{4}\)

There is an error in this calculation; let's revisit the correct method.

From a refined calculation, considering the radius and density impact more accurately within the root, as:

\(v_p = \sqrt{4 \cdot 2} \cdot v_e = 2\sqrt{2} \cdot v_e\)

The correct ratio should be:

\(\text{Ratio} = \frac{v_e}{v_p} = \frac{1}{2\sqrt{2}}\)

Hence, the correct ratio is \(1 : 2\sqrt{2}\).