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Chapter 6: Problem 803

He period of revolution of planet \(\mathrm{A}\) around the sun is 8 times that of \(\mathrm{B}\). The distance of A from the sun is how many times greater than that of \(\mathrm{B}\) from the sun. (A) 2 (B) 3 (C) 4 (D) 5

Short Answer

Expert verified
The distance of A from the Sun is 4 times greater than that of B from the Sun. The correct answer is (C) 4.

Step by step solution

01

Write down the given information

We know that the period of revolution of planet A is 8 times that of planet B. We can represent this relationship mathematically as: \(T_A = 8 T_B\)
02

Apply Kepler's Third Law

According to Kepler's Third Law, the ratio of the squares of the periods of two planets (A and B) is equal to the ratio of the cubes of the semi-major axes of their orbits (represented by a). We can write this relationship as: \(\frac{T_A^2}{T_B^2} = \frac{a_A^3}{a_B^3}\)
03

Substitute the given relationship

Now, we can replace \(T_A\) with \(8 T_B\) in the equation as per the given information. \(\frac{(8 T_B)^2}{T_B^2} = \frac{a_A^3}{a_B^3}\)
04

Simplify the equation

Simplify the left-hand side of the equation to find the ratio of their distances from the Sun in terms of their periods of revolution. \(\frac{(8 T_B)^2}{T_B^2} = \frac{64 T_B^2}{T_B^2} = 64\) So, \(64 = \frac{a_A^3}{a_B^3}\)
05

Determine the relationship between the distances

Now we need to find the cube root of 64, which will give us the factor by which the distance of A is greater than the distance of B from the Sun. \(\sqrt[3]{64} = 4\) Hence, the distance of A from the Sun is 4 times greater than that of B from the Sun. The correct answer is (C) 4.

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Most popular questions from this chapter

A satellite with K.E. \(E_{\mathrm{k}}\) is revolving round the earth in a circular orbit. How much more K.E. should be given to it so that it may just escape into outer space ? (A) \(\mathrm{E}_{\mathrm{k}}\) (B) \(2 \mathrm{E}_{\mathrm{k}}\) (C) \((1 / 2) \mathrm{E}_{\mathrm{k}}\) (D) \(3 \mathrm{E}_{\mathrm{k}}\)

The escape velocity from the earth is about \(11 \mathrm{kms}^{-1}\). The escape velocity from a planet having twice the radius and the same mean density as the earth is \(=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}\). (A) 22 (B) 11 (C) \(5.5\) (D) \(15.5\)

The escape velocity of a projectile from the earth is approximately (A) \(11.2 \mathrm{kms}^{-1}\) (B) \(112 \mathrm{kms}^{-1}\) (C) \(11.2 \mathrm{~ms}^{-1}\) (D) \(1120 \mathrm{kms}^{-1}\)

The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of earth. If the radius of the earth is \(\mathrm{R}\), the radius of planet would be (A) \(2 \mathrm{R}\) (B) \(4 \mathrm{R}\) (C) \(1 / 4 \mathrm{R}\) (D) \(\mathrm{R} / 2\)

According to keplar, the period of revolution of a planet ( \(\mathrm{T}\) ) and its mean distance from the sun (r) are related by the equation (A) \(\mathrm{T}^{3} \mathrm{r}^{3}=\) constant (B) \(\mathrm{T}^{2} \mathrm{r}^{-3}=\) constant (C) \(\mathrm{Tr}^{3}=\) constant (D) \(\mathrm{T}^{2} \mathrm{r}=\) constant
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