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Chapter 6: Problem 731

The escape velocity for a sphere of mass \(\mathrm{m}\) from earth having mass \(\mathrm{M}\) and Radius \(\mathrm{R}\) mass is given by (A) \(\sqrt{[}(2 \mathrm{GM}) / \mathrm{R}]\) (B) \(2 \sqrt{(\mathrm{GM} / \mathrm{R})}\) (C) \(\sqrt{[}(2 \mathrm{GMm}) / \mathrm{R}]\) (D) \(\sqrt{(\mathrm{GM} / \mathrm{R})}\)

Short Answer

Expert verified
The correct escape velocity formula for a sphere of mass \(m\) from Earth having mass \(M\) and radius \(R\) is (A) \(\sqrt{(2 \mathrm{GM}) / \mathrm{R}}\).

Step by step solution

01

Remember the definition of escape velocity

The escape velocity is the minimum speed an object must have to escape from the gravitational force of a planet or any celestial body. In this case, we are considering the object escaping from Earth's gravitational force.
02

Apply conservation of energy principle

Conservation of energy principle states that the total mechanical energy (potential + kinetic) of the system remains constant. In this situation, the total mechanical energy (E) can be represented as: \[E = KE + PE\] Where KE is the kinetic energy and PE is the gravitational potential energy.
03

Formulate Kinetic and Potential energy expressions

The Kinetic energy (KE) is given by: \[KE = \frac{1}{2}mv^2\] Where m is the mass of the object and v is its velocity. The gravitational potential energy (PE) is given by: \[PE = -\frac{GMm}{r}\] Where G is the gravitational constant, M is the mass of Earth, m is the mass of the object, and r is the distance between the center of masses of both the object and Earth.
04

Setting the total mechanical energy to zero and solving for escape velocity

Since the object is escaping the gravitational force, we can equate the total mechanical energy to zero. This will result in the following equation: \[E = KE + PE = 0\] Now, substitute the expressions for KE and PE from Step 3 into this equation: \[\frac{1}{2}mv^2 -\frac{GMm}{r} = 0\]
05

Isolate the velocity in the equation

Now, we need to solve this equation for the escape velocity v. First, we can divide both sides of the equation by m: \[\frac{1}{2}v^2 -\frac{GM}{r} = 0\] Next, add \(\frac{GM}{r}\) to both sides: \[\frac{1}{2}v^2 = \frac{GM}{r}\] Now, multiply both sides by 2: \[v^2 = \frac{2GM}{r}\] Finally, take the square root: \[v = \sqrt{\frac{2GM}{r}}\] Comparing this escape velocity formula to the options given in the exercise, we can identify the correct answer: (A) \(\sqrt{[}(2 \mathrm{GM}) / \mathrm{R}]\)

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Most popular questions from this chapter

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) \(\mathrm{R} / 8\)

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) \(-[(4 \mathrm{GM}) / \mathrm{r}]\)

A particle of mass \(\mathrm{M}\) is situated at the center of a spherical shell of same mass and radius a the magnitude of gravitational potential at a point situated at (a/2) distance from the center will be (A) \([(4 \mathrm{GM}) / \mathrm{a}]\) (B) \((\mathrm{GM} / \mathrm{a})\) (C) \([(2 \mathrm{GM}) / \mathrm{a}]\) (D) \([(3 \mathrm{GM}) / \mathrm{a}]\)

The radii of two planets are respectively \(\mathrm{R}_{1}\) and \(\mathrm{R}_{2}\) and their densities are respectively \(\rho_{1}\) and \(\rho_{2}\) the ratio of the accelerations due to gravity at their surface is (A) $g_{1}: g_{2}=\left(\rho_{1} / R_{1}^{2}\right) \cdot\left(\rho_{2} / R_{2}^{2}\right)$ (B) $\mathrm{g}_{1}: \mathrm{g}_{2}=\mathrm{R}_{1} \mathrm{R}_{2}: \rho_{1} \rho_{2}$ (C) \(g_{1}: g_{2}=R_{1} \rho_{2} \cdot R_{2} p_{1}\) (D) \(g_{1}: g_{2}=R_{1} \rho_{1}: R_{2} \rho_{2}\)

At what distance from the center of earth, the value of acceleration due to gravity \(g\) will be half that of the surfaces \((\mathrm{R}=\) Radius of earth \()\) (A) \(2 \mathrm{R}\) (B) \(\mathrm{R}\) (C) \(1.414 \mathrm{R}\) (D) \(0.414 \mathrm{R}\)
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