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Chapter 7: Problem 61

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 \(2 \mathrm{~km} / \mathrm{s}\).

Short Answer

Expert verified
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 by step solution

01

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.
02

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.
03

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.

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Most popular questions from this chapter

A particle is placed in a field characterized by a value of gravitational potential given by \(V=-k x y\), where \(k\) is a constant. If \(\vec{E}_{g}\) is the gravitational field then, (A) \(\vec{E}_{g}=k(x \hat{i}+y \hat{j})\) and is conservative in nature. (B) \(\vec{E}_{g}=k(y \hat{i}+x \hat{j})\) and is conservative in nature. (C) \(\vec{E}_{g}=k(x \hat{i}+y \hat{j})\) and is non-conservative in nature (D) \(\vec{E}_{g}=k(y \hat{i}+x \hat{j})\) and is non-conservative in nature.

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Three planets of same density have radii \(R_{1}, R_{2}\) and \(R_{3}\) such that \(R_{1}=2 R_{2}=3 R_{3}\). The gravitational field at their respective surfaces are \(g_{1}, g_{2}\) and \(g_{3}\) and escape velocities from their surfaces are \(v_{1}, v_{2}\) and \(v_{3}\) respectively, then (A) \(g_{1} / g_{2}=2\) (B) \(g_{1} / g_{3}=3\) (C) \(v_{1} / v_{2}=1 / 4\) (D) \(v_{1} / v_{3}=3\)

The escape velocity for a body projected vertically upwards from the surface of earth is \(11 \mathrm{~km} / \mathrm{s}\). If the body is projected at an angle of \(45^{\circ}\) with the vertical, the escape velocity will be (A) \(11 \sqrt{2} \mathrm{~km} / \mathrm{s}\) (B) \(22 \mathrm{~km} / \mathrm{s}\) (C) \(11 \mathrm{~km} / \mathrm{s}\) (D) \(\frac{11}{\sqrt{2}} \mathrm{~m} / \mathrm{s}\)
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