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Chapter 6: Problem 749

The acceleration due to gravity on a planet is same as that on earth and its radius is four times that of earth. What will be the value of escape velocity on that planet if it is \(\mathrm{V}_{\mathrm{e}}\) on the earth (A) \(\mathrm{V}_{\mathrm{e}}\) (B) \(2 \mathrm{~V}_{\mathrm{e}}\) (C) \(4 \mathrm{~V}_{\mathrm{e}}\) (D) \(\mathrm{V}_{\mathrm{e}} / 2\)

Short Answer

Expert verified
The escape velocity of the given planet is half of the escape velocity of Earth. Thus, the answer is (D) \(V_{e_{planet}} = \frac{V_{e_{Earth}}}{2}\).

Step by step solution

01

List the given information

The acceleration due to gravity on the planet is same as that on Earth: \(g_{planet} = g_{Earth}\) The radius of the planet is four times the radius of Earth: \(R_{planet} = 4 * R_{Earth}\)
02

Write down the escape velocity formula

The escape velocity (\(V_e\)) is calculated using the formula: \[V_e = \sqrt{\frac{2GM}{R}}\] where - \(V_e\) is the escape velocity - \(G\) is the gravitational constant - \(M\) is the mass of the planet - \(R\) is the radius of the planet
03

Express the mass of the planet in terms of Earth

The mass of the planet can be expressed in terms of Earth using the provided information about the gravity and radius of the planet. Since acceleration due to gravity on both planets is equal: \[g_{Earth} = g_{planet}\] Also, recall that \[g = \frac{GM}{R^2}\] So, for Earth and the given planet: \[\frac{G M_{Earth}}{R_{Earth}^2} = \frac{G M_{planet}}{R_{planet}^2}\] We know that \(R_{planet} = 4 R_{Earth}\). So, we can rewrite the equation: \[\frac{M_{Earth}}{R_{Earth}^2} = \frac{M_{planet}}{(4R_{Earth})^2}\] Solve the equation for the mass of the planet (\(M_{planet}\)): \[M_{planet} = \frac{1}{16} M_{Earth}\]
04

Calculate the escape velocity of the given planet

Now we have enough information to calculate the escape velocity of the given planet. Substitute the values of \(M_{planet}\) and \(R_{planet}\) into the escape velocity formula: \[V_{e_{planet}} = \sqrt{\frac{2G}{16}M_{Earth} \div 4R_{Earth}}\] Simplifying the expression, we get: \[V_{e_{planet}} = \sqrt{\frac{G M_{Earth}}{32R_{Earth}}}\]
05

Compare the escape velocity to Earth's escape velocity

Now, let's find the escape velocity of Earth: \[V_{e_{Earth}} = \sqrt{\frac{2GM_{Earth}}{R_{Earth}}}\] Now, divide \(V_{e_{planet}}\) by \(V_{e_{Earth}}\): \[\frac{V_{e_{planet}}}{V_{e_{Earth}}} = \frac{\sqrt{\frac{G M_{Earth}}{32R_{Earth}}}}{\sqrt{\frac{2GM_{Earth}}{R_{Earth}}}}\] Simplifying the expression, we find: \[\frac{V_{e_{planet}}}{V_{e_{Earth}}} = \frac{1}{2}\] Which indicates that the escape velocity of the given planet is half of the escape velocity of Earth. Thus, the answer is: \[\boxed{(D)~ V_{e_{planet}} = \frac{V_{e_{Earth}}}{2}}.\]

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