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Chapter 6: Problem 796

If the earth is at one- fourth of its present distance from the sun the duration of year will be (A) half the present Year (B) one-eight the present year (C) one-fourth the present year (D) one-sixth the present year

Short Answer

Expert verified
The duration of a year, when the Earth is at one-fourth of its present distance from the sun, is one-eighth the present year. This corresponds to option (B).

Step by step solution

01

Consider Kepler's Third Law

Kepler's Third Law states that the square of the orbital period of a planet (in this case, Earth) is directly proportional to the cube of the semi-major axis (average distance from the sun) of its orbit. Mathematically, this can be expressed as: \(T^2 \propto a^3\)
02

Find the ratio of orbital periods

Let's denote the present orbital period of the Earth as \(T_1\) and its present distance from the sun as \(a_1\). Similarly, let the orbital period when the Earth is at one-fourth of its present distance be \(T_2\) and the distance from sun be \(a_2\). Since the Earth is at one-fourth of its present distance, we have \(a_2 = \frac{1}{4}a_1\). Now we need to find the ratio \( \frac{T_2}{T_1}\)
03

Apply Kepler's Third Law to find the ratio

We can write the ratio of the squared orbital periods as: \(\frac{T_2^2}{T_1^2} = \frac{a_2^3}{a_1^3}\) Since \(a_2 = \frac{1}{4}a_1\), we can substitute this value, and we get: \(\frac{T_2^2}{T_1^2} = \frac{(\frac{1}{4}a_1)^3}{a_1^3}\)
04

Simplify the equation and solve for the ratio

Simplify the above equation as: \(\frac{T_2^2}{T_1^2} = \frac{1}{64}\) Now we can take the square root of both sides to find the ratio of the orbital periods: \(\frac{T_2}{T_1} = \frac{1}{8}\)
05

Compare the ratio with the given options

The ratio \(\frac{T_2}{T_1} = \frac{1}{8}\) indicates that the duration of a year, when the Earth is at one-fourth of its present distance from the sun, is one-eighth of the present year. This corresponds to option (B): (B) one-eighth the present year.

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Most popular questions from this chapter

An artificial satellite moving in a circular orbit around earth has a total (kinetic + potential energy) \(E_{0}\), its potential energy is (A) \(-\mathrm{E}_{0}\) (B) \(1.5 \mathrm{E}_{0}\) (C) \(2 \mathrm{E}_{0}\) (D) \(\mathrm{E}_{0}\)

The additional K.E. to be provided to a satellite of mass \(\mathrm{m}\) revolving around a planet of mass \(\mathrm{M}\), to transfer it from a circular orbit of radius \(\mathrm{R}_{1}\) to another radius \(\mathrm{R}_{2}\left(\mathrm{R}_{2}>\mathrm{R}_{1}\right)\) is (A) $\operatorname{GMm}\left[\left(1 / R_{1}^{2}\right)-\left(1 / R_{2}^{2}\right)\right]$ \(\operatorname{GMm}\left[\left(1 / R_{1}\right)-\left(1 / R_{2}\right)\right]\) (C) $2 \mathrm{GMm}\left[\left(1 / \mathrm{R}_{1}\right)-\left(1 / \mathrm{R}_{2}\right)\right]$ (D) $(1 / 2) \mathrm{GMm}\left[\left(1 / \mathrm{R}_{1}\right)-\left(1 / \mathrm{R}_{2}\right)\right]$

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) \(\mathrm{R} / 2\)

Two small and heavy sphere, each of mass \(\mathrm{M}\), are placed distance r apart on a horizontal surface the gravitational potential at a mid point on the line joining the center of spheres is (A) zero (B) \(-(\mathrm{GM} / \mathrm{r})\) (C) \(-[(2 \mathrm{GM}) / \mathrm{r}]\) (D) \(-[(4 \mathrm{GM}) / \mathrm{r}]\)

If \(\mathrm{V}_{\mathrm{e}}\) and \(\mathrm{V}_{\mathrm{o}}\) are represent the escape velocity and orbital velocity of satellite corresponding to a circular orbit of -adius \(\mathrm{r}\), then A) \(\mathrm{V}_{\mathrm{e}}=\mathrm{V}_{\mathrm{o}}\) (B) \(\sqrt{2} \mathrm{~V}_{\mathrm{o}}=\mathrm{V}_{\mathrm{e}}\) C) \(\mathrm{V}_{\mathrm{e}}=\left(\mathrm{V}_{\mathrm{O}} / \sqrt{2}\right)\) (D) \(\mathrm{V}_{\mathrm{e}}\) and \(\mathrm{V}_{\mathrm{o}}\) are not related
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