Problem 75 - A Complete Resource Book in Physics for JEE Main

Question

A planet in a distance solar system is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is \(11 \mathrm{kms}^{-1}\), the escape velocity from the surface of the planet would be. (A) \(1.1 \mathrm{kms}^{-1}\) (B) \(11 \mathrm{kms}^{-1}\) (C) \(110 \mathrm{kms}^{-1}\) (D) \(0.11 \mathrm{kms}^{-1}\)

Step-by-Step Solution

Answer

The escape velocity from the surface of the planet would be \(110 \mathrm{kms}^{-1}\). Hence, the correct answer is (C) \(110 \mathrm{kms}^{-1}\).

Step 1: Understand the Escape velocity formula

The escape velocity formula is \( v = \sqrt{2GM/R} \), where v is the escape velocity, \(G\) is the gravitational constant, \(M\) is the mass of the body (planet), and \(R\) is the radius of the body(planet).

Step 2: Substitute Earth's escape velocity into the formula

We know the Earth's escape velocity is 11kms^-1. Plugging the values we have, that is the escape velocity v and the Earth's mass (M) and radius (R) where \(M\) is 1 (in earth mass) and \(R\) is 1 (in earth radius), into the formula for escape velocity, we get \(v = \sqrt{2G*1/1}\). So, this gives us the result that \(2G = v^2\) or \(2G = 11^2 = 121\).

Step 3: Calculate the escape velocity of the planet

Now, we calculate the escape velocity for the planet with mass 10 times the Earth's mass and radius 1/10 of the Earth's radius. The new escape velocity \(v'\) is hence \(v' = \sqrt{2G*10/0.1} = \sqrt{(2G)*100} = 10* v\). This gives us \(v' = 10*11 = 110 \mathrm{kms}^{-1}\)