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Chapter 6: Problem 831

A planet is revolving round the sun in elliptical orbit. Velocity at perigee position (nearest) is \(\mathrm{v}_{1} \mid\) and at apogee position (farthest) is \(\mathrm{v}_{2}\) Both these velocities are perpendicular to the joining center of sun and planet \(r\) is the minimum distance and \(\mathrm{r}_{2}\) the maximum distance. (1) when the planet is at perigee position, it wants to revolve in a circular orbit by itself. For this value of \(\mathrm{G}\) (A) Should increase (B) Should decrease (C) data is insufficient (D) will not depend on the value of \(\mathrm{G}\) (2) At apogee position suppose speed of planer is slightly decreased from \(\mathrm{v}_{2}\), then what will happen to minimum distance \(r_{1}\) in the subsequent motion (A) \(r_{1}\) and \(r_{2}\) both will decreases (B) \(\mathrm{r}_{1}\) and \(\mathrm{r}_{2}\) both will increases (C) \(\mathrm{r}_{2}\) will remain as it is while \(\mathrm{r}_{1}\) will increase (D) \(\mathrm{r}_{2}\) will remain as it is while \(\mathrm{r}_{1}\) will decrease

Short Answer

Expert verified
(A) G should increase. (C) r2 will remain as it is while r1 will increase.

Step by step solution

01

Gravitational and Centripetal Force Relationship at Perigee Position

The gravitational force F between the planet and the sun is given by: \[F = G\frac{Mm}{r^2}\] where G is the gravitational constant, M is the mass of the sun, m is the mass of the planet, and r is the perigee distance. At perigee, the centripetal force required for circular motion is given by: \[F_c = \frac{mv_1^2}{r}\] From the force balance, we have: \[G\frac{Mm}{r^2} = \frac{mv_1^2}{r}\]
02

Determine the Dependence on G at the Perigee Position

From the equation derived in step 1, we can solve for G as follows: \[G = \frac{v_1^2}{M}\] Since the speed v1, and mass M are constants and the value of G is directly proportional to v1², we can conclude that for the planet to revolve in a circular orbit by itself at perigee: (A) G should increase.
03

Analyze the Change in Minimum Distance when Speed at Apogee is Decreased

When the planet's speed at apogee position v2 is slightly decreased, it will not have enough centripetal force to maintain the same r2 (maximum distance). As a result, the planet will be pulled closer to the sun, causing r1 (minimum distance) to increase. From this analysis, we deduce that when the speed at the apogee position is decreased: (C) r2 will remain as it is while r1 will increase.

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Most popular questions from this chapter

The escape velocity of an object from the earth depends upon the mass of earth (M), its mean density ( \(p\) ), its radius (R) and gravitational constant (G), thus the formula for escape velocity is (A) \(U=\mathrm{R} \sqrt{[}(8 \pi / 3) \mathrm{Gp}]\) (C) \(\mathrm{U}=\sqrt{(2 \mathrm{GMR})}\) (D) \(U=\sqrt{\left[(2 \mathrm{GMR}) / \mathrm{R}^{2}\right]}\)

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Two bodies of masses \(m_{1}\) and \(m_{2}\) are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual a gravitational attraction Their relative velocity of approach at separation distance \(\mathrm{r}\) between them is (A) $\left[\left\\{2 \mathrm{G}\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right)\right\\} / \mathrm{r}\right]^{-1 / 2}$ (B) $\left[\left\\{2 \mathrm{G}\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)\right\\} / \mathrm{r}\right]^{1 / 2}$ (C) $\left[\mathrm{r} /\left\\{2 \mathrm{G}\left(\mathrm{m}_{1} \mathrm{~m}_{2}\right)\right\\} / \mathrm{r}\right]^{1 / 2}$ (D) $\left[\left(2 \mathrm{Gm}_{1} \mathrm{~m}_{2}\right) / \mathrm{r}\right]^{1 / 2}$

The time period \(\mathrm{T}\) of the moon of planet Mars \((\mathrm{Mm})\) is related to its orbital radius \(\mathrm{R}\) as \((\mathrm{G}=\) Gravitational constant \()\) (A) $\mathrm{T}^{2}=\left[\left(4 \pi^{2} \mathrm{R}^{3}\right) /(\mathrm{GMm})\right]$ (B) $\mathrm{T}^{2}=\left[\left(4 \pi^{2} \mathrm{GR}^{3}\right) /(\mathrm{Mm})\right]$ (C) \(T^{2}=\left[\left(2 \pi R^{2} G\right) /(M m)\right]\) (D) \(\mathrm{T}^{2}=4 \pi \mathrm{Mm} \mathrm{GR}^{2}\)

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