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Chapter 6: Problem 741

The escape velocity of a planet having mass 6 times and radius 2 times as that of earth is (A) \(\sqrt{3} \mathrm{~V}_{\mathrm{e}}\) (B) \(3 \mathrm{~V}_{\mathrm{e}}\) (C) \(\sqrt{2} \mathrm{~V}_{\mathrm{e}}\) (D) \(2 \mathrm{~V}_{\mathrm{e}}\)

Short Answer

Expert verified
The escape velocity of a planet with a mass 6 times and a radius 2 times greater than Earth's is (A) \(\sqrt{3} V_{e}\).

Step by step solution

01

Understand the Escape Velocity Formula

Escape velocity is the minimum speed an object needs to break free from a celestial body's gravitational pull. The escape velocity formula is given by: \[v_{esc} = \sqrt{\frac{2GM}{r}}\] where \(v_{esc}\) is the escape velocity, \(G\) is the gravitational constant, \(M\) is the mass of the celestial body, and \(r\) is its radius.
02

Find the Escape Velocity of Earth

We don't have the values of the gravitational constant, mass of the Earth, and Earth's radius. However, we don't need them since we only need to find the ratio between the other planet's escape velocity and Earth's escape velocity. Let's denote Earth's escape velocity as \(V_{e}\).
03

Apply the Values of the Given Planet

The given planet has a mass of \(6M\) and a radius of \(2r\) (where M and r are Earth's mass and radius). We can plug these values into the formula for escape velocity: \[v_{esc}^{planet} = \sqrt{\frac{2G(6M)}{2r}}\]
04

Simplify the Equation

Now, we can simplify the equation: \[v_{esc}^{planet} = \sqrt{\frac{12GM}{2r}} = \sqrt{\frac{6GM}{r}}\]
05

Calculate the Ratio of the Given Planet's Escape Velocity to Earth's Escape Velocity

Finally, we need to find the ratio between the planet's escape velocity and Earth's escape velocity: \[\frac{v_{esc}^{planet}}{V_{e}} = \frac{\sqrt{\frac{6GM}{r}}}{\sqrt{\frac{2GM}{r}}}\] The mass \(M\) and radius \(r\) are common in both the numerator and the denominator, thus will cancel out. We are left with: \[\frac{v_{esc}^{planet}}{V_{e}} = \sqrt{\frac{6}{2}} = \sqrt{3}\] The answer is (A) \(\sqrt{3} V_{e}\).

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