Problem 61 - A Complete Resource Book in Physics for JEE Main

Question

Calculate the ratio \(m_{0} / m\) for a rocket if it is to escape from the earth. Given escape velocity \(=11.2 \mathrm{~km} / \mathrm{s}\) and exhaust speed of gases is \(2 \mathrm{~km} / \mathrm{s}\).

Step-by-Step Solution

Answer

Using the formula, the ratio of the initial to final mass of the rocket to escape from the earth is calculated as \(m_{0} / m = e^{(v_{e} / v_{g})}\).

Step 1: Identify Known Variables

Identify and list down the known variables. The escape velocity \(v_{e}\) is given as 11.2 km/s and exhaust speed of gases \(v_{g}\) is provided as 2 km/s.

Step 2: Use the Tsiolkovsky rocket equation

Tsiolkovsky rocket equation (or rocket equation), describes the motion of vehicles that follow the basic principle of a rocket: a device that can apply acceleration to itself using thrust by expelling part of its mass with high speed in the opposite direction, due to the conservation of momentum. The equation is given as \(v_{e} = v_{g} * ln(m_{0} / m)\) where \(ln(m_{0} / m)\) is the natural logarithm of the ratio of the initial mass \(m_0\) to the final mass \(m\) of the rocket.

Step 3: Calculate the ratio \(m_{0} / m\)

Rearrange the rocket equation to find the ratio \(m_{0} / m = e^{(v_{e} / v_{g})}\). Substitute the given values into the equation and solve. Make sure that the two velocities are in the same units. Here, both velocities are in km/s, so they are in consistent units.