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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

A satellite is moving around the earth with speed \(\mathrm{v}\) in a circular orbit of radius \(\mathrm{r}\). If the orbit radius is decreased by \(1 \%\) its speed will (A) increase by \(1 \%\) (B) increase by \(0.5 \%\) (C) decreased by \(1 \%\) (C) Decreased by \(0.5 \%\)

The escape velocity of an object from the earth depends upon the mass of earth (M), its mean density ( \(p\) ), its radius (R) and gravitational constant (G), thus the formula for escape velocity is (A) \(U=\mathrm{R} \sqrt{[}(8 \pi / 3) \mathrm{Gp}]\) (C) \(\mathrm{U}=\sqrt{(2 \mathrm{GMR})}\) (D) \(U=\sqrt{\left[(2 \mathrm{GMR}) / \mathrm{R}^{2}\right]}\)

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

A geostationary satellite is orbiting the earth at a height of \(5 \mathrm{R}\) above that of surface of the earth. \(\mathrm{R}\) being the radius of the earth. The time period of another satellite in hours at a height of $2 \mathrm{R}\( from the surface of earth is \)\ldots \ldots \ldots .$ hr (A) 5 (B) 10 (C) \(6 \sqrt{2}\) (D) \(6 / \sqrt{2}\)

Two small and heavy sphere, each of mass \(\mathrm{M}\), are placed distance r apart on a horizontal surface the gravitational potential at a mid point on the line joining the center of spheres is (A) zero (B) \(-(\mathrm{GM} / \mathrm{r})\) (C) \(-[(2 \mathrm{GM}) / \mathrm{r}]\) (D) \(-[(4 \mathrm{GM}) / \mathrm{r}]\)
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