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Chapter 6: Problem 727

A body is projected vertically upwards from the surtace of a planet of radius \(\mathrm{R}\) with a velocity equal to half the escape velocity for that planet. The maximum height attained by the body is $\ldots \ldots \ldots \ldots$ (A) \((\mathrm{R} / 3)\) (B) \((\mathrm{R} / 2)\) (C) \((\mathrm{R} / 4)\) (D) \((\mathrm{R} / 5)\)

Short Answer

Expert verified
The maximum height attained by the body is \(H = R\), which corresponds to option (B).

Step by step solution

01

Write down the known parameters

The radius of the planet is R. The initial velocity (v0) is equal to half the escape velocity, which is given by the formula: \(v_e = \sqrt{\frac{2GM}{R}}\), where G is the gravitational constant, and M is the mass of the planet. So, the initial velocity v0 is: \(v_0 = \frac{1}{2}v_e = \frac{1}{2}\sqrt{\frac{2GM}{R}}\)
02

Apply the conservation of mechanical energy

We can write the total mechanical energy of the system (kinetic energy + gravitational potential energy) at launch (initial) and at the maximum height (final). Initially, the projectile has kinetic energy and potential energy due to its position on the planet's surface: \[E_{initial} = \frac{1}{2}mv_0^2 - \frac{GMm}{R}\] At the maximum height (H), the projectile's velocity is 0 (it momentarily stops moving upwards), so its kinetic energy is 0. However, it has potential energy due to its position at height H: \[E_{final} = -\frac{GMm}{R+H}\] Since the mechanical energy is conserved, we can write: \[E_{initial} = E_{final}\]
03

Solve for the maximum height H

From the energy conservation equation, we get: \[\frac{1}{2}mv_0^2 - \frac{GMm}{R} = -\frac{GMm}{R+H}\] Substitute the initial velocity v0 into the equation and solve for the unknown height H: \[\frac{1}{2}m\left(\frac{1}{2}\sqrt{\frac{2GM}{R}}\right)^2 - \frac{GMm}{R} = -\frac{GMm}{R+H}\] Simplify and solve for H: \[\frac{1}{4}\frac{2GMm}{R} - \frac{GMm}{R} = -\frac{GMm}{R+H}\] \[-\frac{1}{2}\frac{GMm}{R} = -\frac{GMm}{R+H}\] Cancel out the negative signs and GMm, and cross-multiply: \[\frac{1}{2R}(R+H)=1\] Solve for H: \[H = R\left(\frac{1}{\frac{1}{2}} - 1\right) = R\left(2-1\right) = R\] Therefore, the maximum height attained by the body is \(H = R\), which corresponds to option (B).

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Most popular questions from this chapter

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: Even when orbit of a satellite is elliptical, its plane of rotation passes through the center of earth Reason: This is in accordance with the principle of conservation of angular momentum (a) \(\mathrm{A}\) (b) B (c) \(\mathrm{C}\) (d) \(\mathrm{D}\)

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

The additional K.E. to be provided to a satellite of mass \(\mathrm{m}\) revolving around a planet of mass \(\mathrm{M}\), to transfer it from a circular orbit of radius \(\mathrm{R}_{1}\) to another radius \(\mathrm{R}_{2}\left(\mathrm{R}_{2}>\mathrm{R}_{1}\right)\) is (A) $\operatorname{GMm}\left[\left(1 / R_{1}^{2}\right)-\left(1 / R_{2}^{2}\right)\right]$ \(\operatorname{GMm}\left[\left(1 / R_{1}\right)-\left(1 / R_{2}\right)\right]\) (C) $2 \mathrm{GMm}\left[\left(1 / \mathrm{R}_{1}\right)-\left(1 / \mathrm{R}_{2}\right)\right]$ (D) $(1 / 2) \mathrm{GMm}\left[\left(1 / \mathrm{R}_{1}\right)-\left(1 / \mathrm{R}_{2}\right)\right]$

A particle of mass \(10 \mathrm{~g}\) is kept on the surface of a uniform sphere of mass \(100 \mathrm{~kg}\) and radius \(10 \mathrm{~cm}\). Find the work to be done against the gravitational force between them to take the particle is away from the sphere \(\left(\mathrm{G}=6.67 \times 10^{-11} \mathrm{SI}\right.\) unit \()\) (A) \(6.67 \times 10^{-9} \mathrm{~J}\) (B) \(6.67 \times 10^{-10} \mathrm{~J}\) (C) \(13.34 \times 10^{-10} \mathrm{~J}\) (D) \(3.33 \times 10^{-10} \mathrm{~J}\)

When a satellite going round the earth in a circular orbit of radius \(\mathrm{r}\) and speed \(\mathrm{v}\) loses some of its energy, then \(\mathrm{r}\) and \(\mathrm{v}\) changes as (A) \(r\) and \(v\) both will increase (B) \(\mathrm{r}\) and \(\mathrm{v}\) both will decease (C) \(r\) will decrease and \(\mathrm{v}\) will increase (D) \(\mathrm{r}\) will increase and \(\mathrm{v}\) will decrease
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