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Chapter 7: Problem 66

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) \(\frac{11}{\sqrt{2}} \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The escape velocity for a body projected at an angle of \(45^{\circ}\) with the vertical is the same as the escape velocity for a vertical launch, which is \(11 \mathrm{ km/s}\). Therefore, the correct answer is (C) \(11 \mathrm{ km/s}\).

Step by step solution

01

Recall the escape velocity formula

The escape velocity \(v_e\) is calculated using the formula: \[v_e = \sqrt{\frac{2GM}{R}}\] where: - \(G\) is the gravitational constant, \(6.674 \times 10^{-11} \mathrm{Nm^{2}/kg^{2}}\) - \(M\) is the mass of the Earth, \(5.972 \times 10^{24} \mathrm{kg}\) - \(R\) is the distance from the center of the Earth to the surface. We are given that \(v_e = 11 \mathrm{ km/s}\).
02

Calculate the initial energy of the body

Since the projectile is escaping Earth's surface, the mechanical energy is conserved. Therefore, the initial mechanical energy equals the final mechanical energy. At the Earth's surface: - Gravitational potential energy (PE) is given by \[\mathrm{PE} = \frac{GMm}{R}\] - Kinetic energy (KE) is given by \[\mathrm{KE} = \frac{1}{2}mv^2\] When you combine these two equations, you get the initial mechanical energy, \[E_{initial}=\frac{GMm}{R}+\frac{1}{2}mv^2\]
03

Calculate the mechanical energy for a 45-degree angle

If the body is projected at a \(45^{\circ}\) angle with the vertical, we need to find the velocity \(v_{45}\) such that the final energy in this situation, \(E_{final}\), is the same as \(E_{initial}\). The conservation of mechanical energy at a \(45^{\circ}\) angle can be expressed as \[E_{final}=\frac{GMm}{R}+\frac{1}{2}mv_{45}^2\] Since \(E_{initial} = E_{final}\), we can set the two expressions equal to one another and solve for \(v_{45}\).
04

Solve for \(v_{45}\)

Equating the initial and final mechanical energy expressions and solving for \(v_{45}\) gives the following: \[\frac{GMm}{R}+\frac{1}{2}mv^2=\frac{GMm}{R}+\frac{1}{2}mv_{45}^2\] Both PE terms cancel out. We are left with: \[\frac{1}{2}mv^2=\frac{1}{2}mv_{45}^2\] Cancelling out the \(m\) and \(\frac{1}{2}\) terms gives: \[v^2 = v_{45}^2\] Therefore, \(v_{45}=v\) which means that the escape velocity at \(45^{\circ}\) with the vertical is the same as the escape velocity for a vertical launch.
05

Find the answer among the given options

The escape velocity for a body projected at an angle of \(45^{\circ}\) with the vertical is the same as the escape velocity for a vertical launch, which is: \(11 \mathrm{ km/s}\) or (C) \(11 \mathrm{ km/s}\). So the correct answer is (C).

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