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Chapter 6: Problem 809

If \(\mathrm{r}\) denotes the distance between the sun and the earth, then the angular momentum of the earth around the sun is proportional to (A) \(1^{3 / 2}\) (B) \(\mathrm{r}\) (C) \(r^{1 / 2}\) (D) \(r^{2}\)

Short Answer

Expert verified
The angular momentum of the Earth around the Sun is proportional to (C) \(r^{1/2}\).

Step by step solution

01

Recall the formula for angular momentum

The angular momentum (L) of an object in circular motion is given by the formula: L = mvr where m is the mass of the object, v is its tangential velocity, and r is the distance between the object and the center point around which it is moving (in this case, the Sun).
02

Relating tangential velocity to distance

We know from Kepler's laws of planetary motion that the earth's orbital period (T) is proportional to the distance (r) raised to the power (3/2): T ∝ r^(3/2) Additionally, we know that T = 2πr/v. Combining these two relationships, we get v ∝ r^(-1/2).
03

Substituting the velocity relationship in the angular momentum formula

We substitute the relationship for tangential velocity in terms of r from Step 2 into the angular momentum formula: L = m(r^(-1/2))r L ∝ r^(1/2)
04

Match the proportionality with the given options

Comparing our result with the given options, we see that the angular momentum of the Earth around the Sun is proportional to: (C) \(r^{1/2}\)

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Most popular questions from this chapter

If \(\mathrm{g}\) is the acceleration due to gravity at the earth's surface and \(\mathrm{r}\) is the radius of the earth, the escape velocity for the body to escape out of earth's gravitational field is \(\ldots \ldots \ldots\) (A) \(\mathrm{gr}\) (B) \(\sqrt{(2 \mathrm{gr})}\) (C) \(\mathrm{g} / \mathrm{r}\) (D) \(\mathrm{r} / \mathrm{g}\)

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

If the height of a satellite from the earth is negligible in comparison of the radius of the earth \(\mathrm{R}\), the orbital velocity of the satellite is (A) \(\mathrm{gR}\) (B) \((\mathrm{gR} / 2)\) (C) \(\sqrt{(g} / \mathrm{R})\) (D) \(\sqrt{(g R)}\)

There are two planets, the ratio of radius of two planets is \(\mathrm{k}\) but the acceleration due to gravity of both planets are \(\mathrm{g}\) what will be the ratio of their escape velocity. (A) \((\mathrm{kg})^{1 / 2}\) (B) \((\mathrm{kg})^{-1 / 2}\) (C) \((\mathrm{kg})^{2}\) (D) \((\mathrm{kg})^{-2}\)

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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