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Chapter 6: Problem 805

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 \(60 \mathrm{~ms}^{-1}\), What will be its velocity in \(\mathrm{ms}^{-1}\) when it is farthest? (A) 6 (B) 12 (C) 60 (D) 112

Short Answer

Expert verified
The velocity of the comet when it is farthest from the sun is \(12 \mathrm{ms}^{-1}\), which corresponds to answer choice (B).

Step by step solution

01

Write down the given values

We have the maximum distance (farthest) of the comet from the sun, \(r_{max} = 8\times10^{12} \mathrm{m}\), the minimum distance (nearest) of the comet from the sun, \(r_{min} = 1.6\times10^{12} \mathrm{m}\), and its velocity when nearest to the sun, \(v_{min} = 60 \mathrm{ms}^{-1}\). We need to find its velocity when farthest to the sun, \(v_{max}\).
02

Use the conservation of angular momentum

Angular momentum is conserved in this scenario, which means the product of the comet's mass, velocity, and distance from the sun is constant. This can be written as: \(m r_{max} v_{max} = m r_{min} v_{min}\), where m is the mass of the comet. Note that the mass of the comet remains the same throughout its orbit.
03

Solve for \(v_{max}\)

We need to find \(v_{max}\), so we can rewrite the previous equation as: \(v_{max} = \frac{m r_{min} v_{min}}{m r_{max}}\) Since the mass of the comet cancels out, we have: \(v_{max} = \frac{r_{min} v_{min}}{r_{max}}\) Now, plug in the given values to find \(v_{max}\): \(v_{max} = \frac{(1.6\times10^{12} \mathrm{m})(60 \mathrm{ms}^{-1})}{8\times10^{12} \mathrm{m}}\)
04

Calculate the value of \(v_{max}\)

By doing the calculations, we get: \(v_{max} = \frac{96\times10^{12} \mathrm{m^2s^{-1}}}{8\times10^{12} \mathrm{m}} = 12 \mathrm{ms}^{-1}\) The velocity of the comet when it is farthest from the sun is \( 12 \mathrm{ms}^{-1}\), which corresponds to answer choice (B).

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Most popular questions from this chapter

The distance of a geo-stationary satellite from the center of the earth (Radius \(\mathrm{R}=6400 \mathrm{~km}\) ) is nearest to (A) \(5 \mathrm{R}\) (B) \(7 \mathrm{R}\) (C) \(10 \mathrm{R}\) (D) \(18 \mathrm{R}\)

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Two satellites \(\mathrm{A}\) and \(\mathrm{B}\) go round a planet \(\mathrm{p}\) in circular orbits having radii \(4 \mathrm{R}\) and \(\mathrm{R}\) respectively if the speed of the satellite \(\mathrm{A}\) is \(3 \mathrm{~V}\), the speed if satellite \(\mathrm{B}\) will be (A) \(12 \mathrm{~V}\) (B) \(6 \mathrm{~V}\) (C) \(4 / 3 \mathrm{~V}\) (D) \(3 / 2 \mathrm{~V}\)

A satellite of mass \(\mathrm{m}\) is circulating around the earth with constant angular velocity. If radius of the orbit is \(\mathrm{R}_{0}\) and mass of earth \(\mathrm{M}\), the angular momentum about the center of earth is (A) \(m \sqrt{\left(G M R_{0}\right)}\) (B) \(\mathrm{M} \sqrt{\left(\mathrm{GMR}_{\mathrm{o}}\right)}\) (C) \(\left.m \sqrt{(G M} / R_{0}\right)\) (D) \(\mathrm{M} \sqrt{\left(\mathrm{GM} / \mathrm{R}_{\mathrm{o}}\right)}\)

let \(\mathrm{V}\) and \(\mathrm{E}\) denote the gravitational potential and gravitational field at a point. Then the match the following \(\begin{array}{ll}\text { Table }-1 & \text { Table }-2\end{array}\) (A) \(\mathrm{E}=0, \mathrm{~V}=0\) (P) At center of spherical shell (B) \(\mathrm{E} \neq 0, \mathrm{~V}=0\) (Q) At center of solid sphere (C) \(\mathrm{V} \neq 0, \mathrm{E}=0\) (R) at centre of circular ring (D) \(\mathrm{V} \neq 0, \mathrm{E} \neq 0\) (S) At centre of two point masses of equal magnitude (T) None
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