Question

The radius of a planet is twice the radius of earth. Both have almost equal average mass-densities. $v_P$ and $v_E$ are escape velocities of the planet and the earth, respectively, then

• A. $v_p = 1.5 \,v_E$
• B. $v_p = 2 \,v_E$
• C. $v_E = 3\, v_p$
• D. $V_E = 1.5\, V_p$

Answer 1

The Correct Option is B

To solve this problem, let's first recap the concept of escape velocity.

The escape velocity (v) from the surface of a planet or celestial body is given by the formula:

v = \sqrt{\frac{2GM}{R}}

where G is the gravitational constant, M is the mass of the planet, and R is the radius of the planet.

Assuming the mass density (\rho) is the same for both the Earth and the planet, the mass (M) of a body is given by:

M = \rho \cdot \frac{4}{3}\pi R^3

For the Earth (E) and the planet (P), this becomes:

• M_E = \rho \cdot \frac{4}{3}\pi R_E^3
• M_P = \rho \cdot \frac{4}{3}\pi R_P^3

Given that R_P = 2R_E, the mass of the planet becomes:

M_P = \rho \cdot \frac{4}{3}\pi (2R_E)^3 = 8 \rho \cdot \frac{4}{3}\pi R_E^3 = 8 M_E

Thus, substituting the expression for mass back into the escape velocity formula:

• Escape velocity for Earth: v_E = \sqrt{\frac{2GM_E}{R_E}}
• Escape velocity for the Planet: v_P = \sqrt{\frac{2GM_P}{R_P}}

Substitute M_P and R_P:

v_P = \sqrt{\frac{2G(8M_E)}{2R_E}} = \sqrt{\frac{16GM_E}{2R_E}} = \sqrt{8} \cdot \sqrt{\frac{2GM_E}{R_E}} = 2\sqrt{2} \cdot v_E

Thus, v_P = 2v_E.

Therefore, the correct option is v_P = 2v_E.