Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Orbital Period
The orbital period is the time it takes for a planet to complete one full orbit around the sun. For Earth, this is typically known as a year, lasting about 365.25 days. The concept of an orbital period is crucial in understanding the regularity and predictability of planetary motion. It helps astronomers describe the periods of different celestial bodies, whether they are planets, moons, or artificial satellites. By knowing the orbital period, scientists can predict where a planet will be in its path around the sun at a given time. This is particularly useful for important calculations in astronomy and space exploration.
Semi-Major Axis
The semi-major axis is a key component in understanding the shape of a planet's orbit. It is essentially half of the longest diameter of the elliptical path that the planet travels along, often referred to as the "average" distance from the sun. Kepler's laws of planetary motion rely heavily on this measurement, as it helps define the size and nature of an orbit. By knowing the semi-major axis, one can use Kepler's Third Law to determine the orbital period. For instance, Earth's semi-major axis is about 93 million miles or 150 million kilometers.
  • The semi-major axis helps determine the energy and angular momentum of the planet in its orbit.
  • Changes in the semi-major axis, as seen in hypothetical exercises, can significantly affect the orbital period, highlighting its importance in planetary science.
Planetary Motion
Planetary motion is the movement of planets around the sun, guided by gravitational forces. Johannes Kepler, through his laws of planetary motion, was able to mathematically describe these movements. Kepler's First Law states that planets move in elliptical orbits with the sun at one focus while the Second Law states that the line joining the planet and the sun sweeps out equal areas in equal times. Kepler's Third Law, which is pivotal in our exercise, highlights the relationship between the square of the orbital period and the cube of the semi-major axis of a planet's orbit.
  • Kepler's laws improved the accuracy of celestial navigation and helped establish the heliocentric view of the solar system.
  • Understanding planetary motion is key for calculating and predicting the positions of planets, which aids in space travel and the study of other celestial phenomena.
Distance from the Sun
The distance between a planet and the sun plays a crucial role in defining various attributes of its orbit, such as speed and duration. For Earth, this distance affects not just the length of a year but also the amount of sunlight and heat we receive, which influences our climate. When considering Kepler's laws, the distance from the sun is directly linked to the semi-major axis of a planet's orbit. This hypothetical exercise shows how drastically changes in this distance can alter the orbital period.
  • If the Earth were closer to the sun, as in this exercise where the distance is one-fourth, the planet would orbit faster and the year would be shorter.
  • The distance from the sun is fundamental in determining a planet's habitability, as well as the conditions of other celestial objects in the solar system.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

orbital velocity of earth's satellite near the surface is \(7 \mathrm{kms}^{-1}\). when the radius of orbit is 4 times that of earth's radius, then orbital velocity in that orbit is \(=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}\) (A) \(3.5\) (B) 17 (C) 14 (D) 35

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) \(\mathrm{r} / \mathrm{g}\)

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) \(\sqrt{(g R)}\)

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

In a gravitational field, at a point where the gravitational potential is zero (A) The gravitational field is necessarily zero (B) The gravitational field is not necessarily zero (C) Nothing can be said definitely, about the gravitational field (D) None of these

See all solutions

Recommended explanations on English Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free