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Chapter 6: Problem 733

The escape velocity from the earth is about \(11 \mathrm{kms}^{-1}\). The escape velocity from a planet having twice the radius and the same mean density as the earth is \(=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}\). (A) 22 (B) 11 (C) \(5.5\) (D) \(15.5\)

Short Answer

Expert verified
The escape velocity of the new planet is 22 km/s.

Step by step solution

01

Recall the escape velocity formula

The formula for escape velocity is given by: \(v_e = \sqrt{\frac{2GM}{r}}\), where \(v_e\) is the escape velocity, \(G\) is the gravitational constant, \(M\) is the mass of the planet, and \(r\) is the radius of the planet.
02

Find the new planet's mass

The volume of a sphere is given by the formula \(V = \frac{4}{3}\pi r^3\). Since the new planet has twice the radius of Earth, its volume will be 8 times larger. Since the new planet has the same mean density as Earth, its mass will also be 8 times larger. Let the mass of Earth be \(M_e\), the mass of the new planet is \(8M_e\).
03

Find the new planet's radius

The new planet's radius is twice the radius of Earth. Let the radius of Earth be \(r_e\), the radius of the new planet is \(2r_e\).
04

Calculate the escape velocity of the new planet

Using the escape velocity formula, substitute the mass and radius of the new planet to find its escape velocity: \(v_{e_{new}} = \sqrt{\frac{2G(8M_e)}{2r_e}}\) Notice that the \(2\) in the numerator and the \(2\) in the denominator will cancel out: \(v_{e_{new}} = \sqrt{\frac{8GM_e}{r_e}}\)
05

Recall the escape velocity of Earth and make the comparison

We know that the escape velocity from Earth is about 11 km/s, so: \(v_{e_{Earth}} = \sqrt{\frac{2GM_e}{r_e}} = 11\) Now, we can square both sides of this equation and obtain: \(\frac{2GM_e}{r_e} = 121\) Now, compare the escape velocity of the new planet with that of Earth: \(\frac{v_{e_{new}}}{v_{e_{Earth}}} = \frac{\sqrt{\frac{8GM_e}{r_e}}}{\sqrt{\frac{2GM_e}{r_e}}} = \sqrt{\frac{8GM_e}{r_e} \cdot \frac{r_e}{2GM_e}} = \sqrt{4} = 2\) This implies that the escape velocity of the new planet is twice that of Earth, or: \(v_{e_{new}} = 2 \cdot 11 = 22 \mathrm{kms}^{-1}\) So the correct answer is (A) 22.

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Most popular questions from this chapter

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) \(44.8\)

Direction (Read the following questions and choose) (A) If both Assertion and Reason are true and the Reason is correct explanation of assertion (B) If both Assertion and Reason are true, but reason is not correct explanation of the Assertion (C) If Assertion is true, but the Reason is false (D) If Assertion is false, but the Reason is true Assertion : Unit of gravitational field intensity is $\mathrm{N} / \mathrm{kg}\( or \)\mathrm{ms}^{-2}\( Reason: Gravitational field intensity \)[$ Force \() /(\) mass $)]=(\mathrm{N} / \mathrm{kg})=\left[\left(\mathrm{kg} \cdot \mathrm{m} / \mathrm{sec}^{2}\right) / \mathrm{kg}\right]=\mathrm{ms}^{-2}$ (a) \(\mathrm{A}\) (b) \(\mathrm{B}\) (c) \(\mathrm{C}\) (d) D

The earth revolves round the sun in one year. If distance between then becomes double the new period will be years. (A) \(0.5\) (B) \(2 \sqrt{2}\) (C) 4 (D) 8

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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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