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Chapter 9: Problem 47

For a satellite escape velocity is \(11 \mathrm{~km} / \mathrm{s}\). If the satellite is launched at an angle of \(60^{\circ}\) with the vertical, then escape velocity will be (a) \(11 \mathrm{~km} / \mathrm{s}\) (b) \(11 \sqrt{3} \mathrm{~km} / \mathrm{s}\) (c) \(\frac{11}{\sqrt{3}} \mathrm{~km} / \mathrm{s}\) (d) \(33 \mathrm{~km} / \mathrm{s}\)

Short Answer

Expert verified
The escape velocity will be \(11 \mathrm{~km} / \mathrm{s}\)

Step by step solution

01

Understanding the concept of escape velocity

The escape velocity of a satellite is the minimum velocity it needs to escape the gravitational field of the earth and it is independent of the direction of launch. Hence, the angle of launch doesn't affect this value.
02

Apply the concept to the problem

Given that the escape velocity is \(11 \mathrm{~km} / \mathrm{s}\) and the satellite is launched at an angle of \(60^{\circ}\), according to our understanding in step 1, the escape velocity remains \(11 \mathrm{~km} / \mathrm{s}\).

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Most popular questions from this chapter

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