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Chapter 6: Problem 746

Escape velocity on the surface of earth is \(11.2 \mathrm{kms}^{-1}\) Escape velocity from a planet whose masses the same as that of earth and radius $1 / 4\( that of earth is \)=\ldots \ldots \ldots \mathrm{kms}^{-1}$ (A) \(2.8\) (B) \(15.6\) (C) \(22.4\) (D) \(44.8\)

Short Answer

Expert verified
The escape velocity of the planet with the same mass as Earth and radius 1/4 that of Earth is \(22.4 \mathrm{kms}^{-1}\) (option C).

Step by step solution

01

Write down the given information

The given information: Escape velocity of Earth (\(v_{Earth}\)) = 11.2 km/s Mass of the planet (\(M\)) = Mass of Earth Radius of the planet (\(R\)) = 1/4 * Radius of Earth
02

Write down the formula for escape velocity

Escape velocity formula: \[v = \sqrt{\frac{2GM}{R}}\]
03

Set up proportion for the planet and Earth

We will set up a proportion to relate the escape velocities of the planet and Earth: \[\frac{v_{planet}}{v_{Earth}} = \sqrt{\frac{R_{Earth}}{R_{planet}}}\]
04

Substitute the given values

Substitute the given values into the proportion: \[\frac{v_{planet}}{11.2} = \sqrt{\frac{4R_{planet}}{R_{planet}}}\]
05

Solve for the escape velocity of the planet

Solve the proportion to find the escape velocity (\(v_{planet}\)): \[v_{planet} = 11.2 * \sqrt{4}\] \[v_{planet} = 11.2 * 2 = 22.4 \mathrm{kms}^{-1}\] So, the escape velocity of the planet is 22.4 km/s, which corresponds to option C.

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Most popular questions from this chapter

The orbital speed of jupiter is (A) greater than the orbital speed of earth (B) less than the orbital speed of earth (C) equal to the orbital speed of earth (D) zero

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) \(\mathrm{V}_{\mathrm{e}} / 2\)

The escape velocity of a projectile from the earth is approximately (A) \(11.2 \mathrm{kms}^{-1}\) (B) \(112 \mathrm{kms}^{-1}\) (C) \(11.2 \mathrm{~ms}^{-1}\) (D) \(1120 \mathrm{kms}^{-1}\)

Two bodies of masses \(m_{1}\) and \(m_{2}\) are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual a gravitational attraction Their relative velocity of approach at separation distance \(\mathrm{r}\) between them is (A) $\left[\left\\{2 \mathrm{G}\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right)\right\\} / \mathrm{r}\right]^{-1 / 2}$ (B) $\left[\left\\{2 \mathrm{G}\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)\right\\} / \mathrm{r}\right]^{1 / 2}$ (C) $\left[\mathrm{r} /\left\\{2 \mathrm{G}\left(\mathrm{m}_{1} \mathrm{~m}_{2}\right)\right\\} / \mathrm{r}\right]^{1 / 2}$ (D) $\left[\left(2 \mathrm{Gm}_{1} \mathrm{~m}_{2}\right) / \mathrm{r}\right]^{1 / 2}$

If the height of a satellite from the earth is negligible in comparison of the radius of the earth \(\mathrm{R}\), the orbital velocity of the satellite is (A) \(\mathrm{gR}\) (B) \((\mathrm{gR} / 2)\) (C) \(\sqrt{(g} / \mathrm{R})\) (D) \(\sqrt{(g R)}\)
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