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Chapter 7: Problem 64

The escape velocity of a body depends upon mass as (A) \(m^{0}\) (B) \(m^{1}\) (C) \(m^{2}\) (D) \(m^{3}\)

Short Answer

Expert verified
The escape velocity depends on the mass of a body as \(M^1\). Therefore, the correct answer is option (B): \(m^1\).

Step by step solution

01

Write down the formula for escape velocity.

The escape velocity formula 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 celestial body, and \(r\) is the distance from its center. In this case, we are concerned with how the escape velocity depends on the mass of the body. We can observe that in the formula, mass is in the numerator and raised to the power of 1.
02

Analyze the relation between escape velocity and mass.

From the escape velocity formula, we can see that the mass of the celestial body, \(M\), has a power of 1 as it is directly multiplied in the formula: \[ v_e = \sqrt{\frac{2G*(M^1)}{r}} \] Thus, the escape velocity depends on mass as \(M^1\).
03

Match the relation with the given options.

We found that the escape velocity depends on the mass of the body as \(M^1\). Now, we will match this with the given options: (A) \(m^0\) (B) \(m^1\) (C) \(m^2\) (D) \(m^3\) As we can see, the correct answer is option (B): \(m^1\).

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Most popular questions from this chapter

If three uniform spheres, each having mass \(M\) and radius \(R\), are kept in such a way that each touches the other two, the magnitude of the gravitational force on any sphere due to the other two is (A) \(\frac{G M^{2}}{4 R^{2}}\) (B) \(\frac{2 G M^{2}}{R^{2}}\) (C) \(\frac{2 G M^{2}}{4 R^{2}}\) (D) \(\frac{\sqrt{3} G M^{2}}{4 R^{2}}\)

Four particles of equal mass \(M\) move along a circle of radius \(R\) under the action of their mutual gravitational attraction. The speed of each particle is (A) \(\frac{G M}{R}\) (B) \(\sqrt{\left(\frac{G M}{R}\right)}\) (C) \(\sqrt{\left[\frac{G M}{R}\left(\frac{2 \sqrt{2}+1}{4}\right)\right]}\) (D) \(\sqrt{\left[\frac{G M}{R}(\sqrt{2}+1)\right]}\)

The period of a satellite in a circular orbit around a planet is independent of, (A) the mass of the planet. (B) the radius of the planet. (C) the mass of the satellite. (D) all of three parameters \(a, b\) and \(c\).

From a solid sphere of mass \(M\) and radius \(R\), a spherical portion of radius \(\frac{R}{2}\) is removed, as shown in Fig. 7.17. Taking gravitational potential \(V=0\) at \(r=\infty\), the potential at the centre of the cavity thus formed is \((G=\) gravitational constant) (A) \(\frac{-G M}{R}\) (B) \(\frac{-2 G M}{3 R}\) (C) \(\frac{-2 G M}{R}\) (D) \(\frac{-G M}{2 R}\)

At a height above the surface of the earth equal to the radius of the earth the value of \(g\) (acceleration due to gravity on the surface of the earth) will be nearly (A) Zero (B) \(\sqrt{g}\) (C) \(\frac{g}{4}\) (D) \(\frac{g}{2}\)
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