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Chapter 6: Problem 732

The escape velocity for a rocket from earth is \(11.2 \mathrm{kms}^{-1}\) value on a planet where acceleration due to gravity is double that on earth and diameter of the planet is twice that of earth will be $=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}$ (A) \(11.2\) (B) \(22.4\) (C) \(5.6\) (C) \(53.6\)

Short Answer

Expert verified
The escape velocity for a rocket from a planet where the acceleration due to gravity is double that on Earth and the diameter of the planet is twice that of Earth is \(22.4 \mathrm{kms^{-1}}\), which corresponds to option (B).

Step by step solution

01

Find the Relationship between Mass and Radius of the Other Planet

Start by setting up the gravitational force equations for both Earth and the other planet: For Earth: \(F_E = G \frac{M_E m}{R_E^2}\), For the other planet: \(F_P = G \frac{M_P m}{R_P^2}\), where \(F_E\) and \(F_P\) are the gravitational forces on Earth and the other planet respectively, \(M_E\) and \(M_P\) are their masses, and \(R_E\) and \(R_P\) are their radii. We are given that the acceleration due to gravity on the other planet is double that on Earth: \(\frac{F_P}{m} = 2 \frac{F_E}{m}\).
02

Solve for the Mass of the Other Planet

Cancel out m and G from both sides: \(\frac{M_P}{R_P^2} = 2 \frac{M_E}{R_E^2}\). We are given that \(R_P = 2 R_E\), so substitute it: \(\frac{M_P}{(2 R_E)^2} = 2 \frac{M_E}{R_E^2}\). Now, solve for \(M_P\): \(M_P = \frac{8}{2} M_E = 4 M_E\).
03

Calculate Escape Velocity of the Other Planet

Now we have the relationship: \(R_P = 2 R_E\) and \(M_P = 4 M_E\). Use the escape velocity formula for the other planet, substitute the relationship, and simplify: \(v_{eP} = \sqrt{\frac{2G(4M_E)}{2R_E}} = \sqrt{\frac{8GM_E}{2R_E}} = \sqrt{4 \frac{2GM_E}{R_E}}\). We know that escape velocity for Earth is \(11.2 \mathrm{kms^{-1}}\), and it can be calculated from the escape velocity formula as: \(11.2 = \sqrt{\frac{2GM_E}{R_E}}\). Now substitute this value in the escape velocity equation for the other planet: \(v_{eP} = \sqrt{4 \times (11.2)^2} = 2 \times 11.2 = 22.4 \mathrm{kms^{-1}}\).
04

Pick the Correct Answer

The correct answer is 22.4 kms⁻¹, which corresponds to option (B).

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Most popular questions from this chapter

The time period \(\mathrm{T}\) of the moon of planet Mars \((\mathrm{Mm})\) is related to its orbital radius \(\mathrm{R}\) as \((\mathrm{G}=\) Gravitational constant \()\) (A) $\mathrm{T}^{2}=\left[\left(4 \pi^{2} \mathrm{R}^{3}\right) /(\mathrm{GMm})\right]$ (B) $\mathrm{T}^{2}=\left[\left(4 \pi^{2} \mathrm{GR}^{3}\right) /(\mathrm{Mm})\right]$ (C) \(T^{2}=\left[\left(2 \pi R^{2} G\right) /(M m)\right]\) (D) \(\mathrm{T}^{2}=4 \pi \mathrm{Mm} \mathrm{GR}^{2}\)

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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) \(44.8\)
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