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Chapter 8: Problem 57

The maximum and minimum distance of a comet from the sun are \(8 \times 10^{12} \mathrm{~m}\) and \(1.6 \times 10^{12} \mathrm{~m}\). If its velocity when nearest to the sun is \(60 \mathrm{~m} / \mathrm{s}\), what will be its velocity in \(\mathrm{m} / \mathrm{s}\) when it is farthest (a) 12 (b) 60 (c) 112 (d) 6

Short Answer

Expert verified
The velocity of the comet when it is farthest from the sun is 12 m/s. Therefore, the correct choice is (a) 12.

Step by step solution

01

Identifying known variables

First identify all the known variables from the exercise. The velocity of the comet when it is nearest to the sun \(v_1 = 60 m/s\). The minimum distance of the comet from the sun or the nearest point \(r_1 = 1.6 x 10^{12} m\). The maximum distance of the comet from the sun or the farthest point \(r_2 = 8 x 10^{12} m\). The velocity of the comet at the farthest point \(v_2\) is what we need to find.
02

Applying the Conservation of Angular Momentum

According to the law of conservation of angular momentum, the angular momentum of the comet when it is nearest to the sun equals the angular momentum when the comet is farthest from the sun. That is \(v_1*r_1 = v_2*r_2\). Now, we can calculate \(v_2\) by rearranging this formula, that gives \(v_2 = (v_1*r_1)/r_2\).
03

Calculation

Substitute the known values into the formula, we get \(v_2 = (60 m/s * 1.6 x 10^{12} m) / (8 x 10^{12} m)\). Solving this we get \(v_2 = 12 m/s\). So, when the comet is farthest from the sun, its velocity is \(12 m/s\).

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Most popular questions from this chapter

The escape velocity from the earth is about \(11 \mathrm{~km} / \mathrm{s}\). The escape velocity from a planet having twice the radius and the same mean density as the earth, is (a) \(22 \mathrm{~km} / \mathrm{s}\) (b) \(11 \mathrm{~km} / \mathrm{s}\) (c) \(5.5 \mathrm{~km} / \mathrm{s}\) (d) \(15.5 \mathrm{~km} / \mathrm{s}\)

The moon's radius is \(1 / 4\) that of the earth and its mass is \(1 / 80\) times that of the earth. If \(g\) represents the acceleration due to gravity on the surface of the earth, that on the surface of the moon is (a) \(\frac{g}{4}\) (b) \(\frac{g}{5}\) (c) \(\frac{g}{6}\) (d) \(\frac{g}{8}\)

Given radius of earth ' \(R\) ' and length of a day 'T' the height of a geostationary satellite is [G - Gravitational constant, \(M\) - Mass of earth] (a) \(\left(\frac{4 \pi^{2} G M}{T^{2}}\right)^{1 / 3}\) (b) \(\left(\frac{4 \pi G M}{R^{2}}\right)^{1 / 3}-R\) (c) \(\left(\frac{G M T^{2}}{4 \pi^{2}}\right)^{1 / 3}-R\) (d) \(\left(\frac{G M T^{2}}{4 \pi^{2}}\right)^{1 / 3}+R\)

A satellite is revolving round the earth in circular orbit at some height above surface of earth. It takes \(5.26 \times 10^{3}\) seconds to complete a revolution while its centripetal acceleration is \(9.92 \mathrm{~m} / \mathrm{s}^{2}\). Height of satellite above surface of earth is (Radius of earth \(6.37 \times 10^{6} \mathrm{~m}\) ) (a) \(70 \mathrm{~km}\) (b) \(120 \mathrm{~km}\) (c) \(170 \mathrm{~km}\) (d) \(220 \mathrm{~km}\)

Two satellites are moving around the earth in circular orbits at height \(R\) and \(3 R\) respectively, \(R\) being the radius of the earth, the ratio of their kinetic energies is (a) 2 (b) 4 (c) 8 (d) 16
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