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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

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A planet moving along an elliptical orbit is closest to the sun at a distance \(\mathrm{r}_{1}\) and farthest away at a distance of \(\mathrm{r}_{2}\). If \(\mathrm{v}_{1}\) and \(\mathrm{v}_{2}\) are the liner velocities at these points respectively, then the ratio \(\left(\mathrm{v}_{1} / \mathrm{v}_{2}\right)\) is \(\ldots \ldots \ldots \ldots\) (A) \(\left(\mathrm{r}_{1} / \mathrm{r}_{2}\right)\) (B) \(\left(\mathrm{r}_{1} / \mathrm{r}_{2}\right)^{2}\) (C) \(\left(\mathrm{r}_{2} / \mathrm{r}_{1}\right)\) (D) \(\left(\mathrm{r}_{2} / \mathrm{r}_{1}\right)^{2}\)
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