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Chapter 6: Problem 744

Given mass of the moon is \((1718)\) of the mass of the earth and corresponding radius is \((1 / 4)\) of the earth, If escape velocity on the earth surface is \(11.2 \mathrm{kms}^{-1}\) the value of same on the surface of moon is $=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}$. (A) \(0.14\) (B) \(0.5\) (C) \(2.5\) (D) 5

Short Answer

Expert verified
The escape velocity on the Moon's surface is approximately 0.84 kms⁻¹, which is closest to option (B) \(0.5\) kms⁻¹.

Step by step solution

01

Write down the given values for mass and radius ratios of Moon

Mass of Moon is \(\frac{1}{718}\) times the mass of Earth (Mₑ): Mₘ = \(\frac{1}{718} Mₑ\) Radius of Moon is \(\frac{1}{4}\) times the radius of Earth (Rₑ): Rₘ = \(\frac{1}{4} Rₑ\) Escape velocity on Earth's surface is: Vₑ = 11.2 kms⁻¹ We need to find escape velocity on Moon's surface (Vₘ).
02

Write down the formula for escape velocity and substitute values for Earth

Escape Velocity = \(\sqrt{\frac{2GM}{R}}\) For escape velocity on Earth's surface (Vₑ): Vₑ = \(\sqrt{\frac{2GMₑ}{Rₑ}}\)
03

Divide escape velocities of Moon and Earth

Divide escape velocities of Moon (Vₘ) and Earth (Vₑ) to find the ratio: \(\frac{Vₘ}{Vₑ}\) = \(\frac{\sqrt{\frac{2G(\frac{1}{718} Mₑ)}{\frac{1}{4} Rₑ}}}{\sqrt{\frac{2GMₑ}{Rₑ}}}\)
04

Simplify the equation and plug in the known values

The gravitational constant (G) and Earth's mass (Mₑ) cancel each other out, simplifying the equation: \(\frac{Vₘ}{Vₑ}\) = \(\sqrt{\frac{\frac{1}{718}}{\frac{1}{4}}}\) Given that the escape velocity on Earth's surface is Vₑ = 11.2 kms⁻¹, plug in the value and solve for Vₘ: Vₘ = Vₑ × \(\sqrt{\frac{\frac{1}{718}}{\frac{1}{4}}}\) = 11.2 × \(\sqrt{\frac{\frac{1}{718}}{\frac{1}{4}}}\)
05

Calculate the escape velocity on Moon and choose the correct option

Now we can find the escape velocity on Moon's surface: Vₘ = 11.2 × \(\sqrt{\frac{4}{718}}\) = 11.2 × \(\sqrt{\frac{1}{179}}\) Vₘ ≈ 0.84 kms⁻¹ From the given options, none of them exactly match 0.84 kms⁻¹. However, the closest option is (B) \(0.5\) kms⁻¹, which is the most accurate option among the given choices.

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Most popular questions from this chapter

The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of earth. If the radius of the earth is \(\mathrm{R}\), the radius of planet would be (A) \(2 \mathrm{R}\) (B) \(4 \mathrm{R}\) (C) \(1 / 4 \mathrm{R}\) (D) \(\mathrm{R} / 2\)

A body of mass \(\mathrm{m} \mathrm{kg}\) starts falling from a point $2 \mathrm{R}\( above the earth's surface. Its \)\mathrm{K} . \mathrm{E}$. when it has fallen to a point ' \(\mathrm{R}\) ' above the Earth's surface $=\ldots \ldots \ldots \ldots J$ [R - Radius of Earth, M-mass of Earth G-Gravitational constant \(]\) (A) \((1 / 2)[(\mathrm{GMm}) / \mathrm{R}]\) (B) \((1 / 6)[(\mathrm{GMm}) / \mathrm{R}]\) (C) \((2 / 3)[(\mathrm{GMm}) / \mathrm{R}]\) (D) \((1 / 3)[(\mathrm{GMm}) / \mathrm{R}]\)

A body weight \(500 \mathrm{~N}\) on the surface of the earth. How much would it weight half way below the surface of earth (A) \(125 \mathrm{~N}\) (B) \(250 \mathrm{~N}\) (C) \(500 \mathrm{~N}\) (D) \(1000 \mathrm{~N}\)

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If the radius of earth is \(\mathrm{R}\) then height \({ }^{\prime} \mathrm{h}\) ' at which value of ' \(\mathrm{g}\) ' becomes one-fourth is (A) \(\mathrm{R} / 4\) (B) \(3 \mathrm{R} / 4\) (C) \(\mathrm{R}\) (D) \(\mathrm{R} / 8\)
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