/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 There have been suggestions that... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

There have been suggestions that the value of the gravitational constant \(G\) becomes smaller when considered over very large time period (in billions of years) in the future. If that happens, for our earth, (a) nothing will change (b) we will become hotter after billions of years (c) we will be going around but not strictly in closed orbits (d) after sufficiently long time we will leave the solar system

Short Answer

Expert verified
(d) After sufficiently long time we will leave the solar system.

Step by step solution

01

Understanding the Problem

The problem is about the implication of a decreasing gravitational constant \(G\) in the future. \(G\) defines the strength of gravity in the universe. If \(G\) becomes smaller, the force of gravity will decrease, which can affect celestial motions.
02

Explore the Effects on Earth's Orbit

Earth's orbit around the sun is determined by the gravitational force \(F\), which is derived from \(F = \frac{Gm_1m_2}{r^2}\), where \(m_1\) and \(m_2\) are masses involved, and \(r\) is the distance between them. If \(G\) decreases, \(F\) will decrease, potentially altering Earth's orbit.
03

Analyzing the Choices

- **Choice (a):** If \(G\) affects gravity significantly over time, it's unlikely that 'nothing will change.'- **Choice (b):** There is no direct link between decreased \(G\) and increased global temperatures.- **Choice (c):** A weaker gravitational pull might prevent Earth from maintaining its current orbit.- **Choice (d):** If \(G\) reduces significantly, Earth's orbit could expand, possibly leading the planet to eventually depart the solar system.
04

Choose the Most Logical Outcome

Given that a decrease in \(G\) weakens the gravitational force, Earth's orbit could destabilize over billions of years, making **Option (d)** the most likely scenario as the Earth could eventually leave the solar system due to insufficient gravitational force to maintain its orbit.

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 91Ó°ÊÓ!

Key Concepts

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

Celestial Mechanics
Celestial mechanics is a branch of astronomy that studies the motions of celestial objects. This field explores how planets, stars, and other astronomical bodies move through space. It is grounded in the principles of classical mechanics and utilizes mathematical methods to predict the positions and movements of these bodies. A key aspect of celestial mechanics is understanding how gravitational forces influence these movements.

The foundational work by Sir Isaac Newton, such as the Law of Universal Gravitation, plays a crucial role in celestial mechanics. This law states that every point mass attracts every other point mass in the universe with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This relationship helps astronomers calculate the orbits of planets and moons, predict celestial events, and study the dynamic evolution of our solar system.

Celestial mechanics is not just theoretical; it's applied in mission planning for spacecraft, guiding satellites, and understanding gravitational interactions between bodies in space. If changes occur in any influencing factors, such as the gravitational constant, it could significantly impact our predictions and calculations in celestial mechanics.
Gravitational Force
Gravitational force is the attractive force exerted by objects that have mass. It is one of the four fundamental forces of nature and is described by the equation: \[ F = \frac{Gm_1m_2}{r^2} \] where:- \( F \) is the gravitational force,- \( G \) is the gravitational constant, - \( m_1 \) and \( m_2 \) are the masses of the objects,- \( r \) is the distance between the centers of the two masses.

This formula indicates that gravitational force increases with the product of the two masses and decreases with the square of the distance between them. Gravitational force is what keeps planets in orbit around stars, moons orbiting planets, and structures like galaxies bound together.

Changes to the gravitational constant \( G \) can alter the strength of gravitational interactions. If \( G \) were to decrease, as was hypothesized in the exercise, the resultant gravitational pull would weaken. For objects in space, like planets and satellites, this means they might drift from their established paths or orbits due to lower gravitational binding forces.
Earth's Orbit
Earth's orbit is the path it follows as it moves around the Sun. It is largely determined by the gravitational force exerted by the Sun, which holds Earth in its elliptical orbit. This orbit keeps Earth at an average distance of about 93 million miles from the Sun.

If the gravitational constant \( G \) were to decrease over time, as suggested in the problem statement, the gravitational pull on Earth would diminish. This could cause Earth's orbit to slowly expand outward. An expanding orbit means a longer revolution period around the Sun and potential changes to Earth's climate and solar exposure.

In extreme scenarios, as described in the solution, Earth could eventually break free from the Sun’s gravitational hold if \( G \) becomes small enough. This would send Earth drifting into deeper space, deviating significantly from its current orbit, and dramatically altering the conditions necessary for life as we know it. This is why understanding the constancy of gravitational parameters is crucial in predicting long-term planetary dynamics.

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

A clock \(S\) is based on oscillation of a spring and a clock \(P\) is based on pendulum motion. Both clock run at the same rate on earth. On a planet having the same density as earth but twice the radius, (a) \(S\) will run faster than \(P\) (b) \(P\) will run faster than \(S\) (c) both will run at the same rate as on the earth (d) both will run at the same rate which will be different from that on the earth

Moment of inertia of cylinder about an axis through the centre and perpendicular to its axis is $$ I_{c}=M\left(\frac{R^{2}}{4}+\frac{L^{2}}{12}\right) $$ Using theorem of parallel axes, moment of inertia of the cylinder about an axis through its edge would be $$ I=I_{c}+M\left(\frac{L}{2}\right)^{2}=M\left(\frac{R^{2}}{4}+\frac{L^{2}}{12}+\frac{L^{2}}{4}\right)=M\left(\frac{R^{2}}{4}+\frac{L^{2}}{3}\right) $$ When \(L=6 R, \quad I_{h}=\frac{49}{4} M R^{2}\)

The period of revolution of planet \(A\) around the sun is 8 times that \(B\). The distance of a from the sun is how many times greater than that of \(B\) from the sun? (a) 2 (b) 3 (c) 4 (d) 5

A uniform ring of mass \(M\) and radius \(r\) is placed directly above a uniform sphere of mass \(8 M\) and of same radius \(R\). The centre of the ring is at a distance of \(d=\sqrt{3} R\) from the centre of the sphere. The gravitational attraction between the sphere and the ring is (a) \(\frac{G M^{2}}{R^{2}}\) (b) \(\frac{3 G M^{2}}{2 R^{2}}\) (c) \(\frac{2 G M^{2}}{\sqrt{2} R^{2}}\) (d) \(\frac{\sqrt{3} G M^{2}}{R^{2}}\)

Angular accelerations is time derivative of angular speed and angular speed is time derivative of angular displacement. By definition \(\alpha=\frac{d \omega}{d t}\) i.e., $$ d \omega=\alpha d t $$ So, if in time \(t\) the angular speed of a body changes from \(\omega_{0}\) to \(\omega\) $$ \int_{\omega_{0}}^{\omega} d \omega=\int_{0}^{t} \alpha d t $$ If \(\alpha\) is constant $$ \omega-\omega_{0}=\alpha t $$ Now, as by definition $$ \omega=\frac{d \theta}{d t} $$ and $$ \theta=\frac{d \omega}{d t} $$ Eq. (i) becomes i.e., $$ \begin{aligned} &\frac{d \theta}{d t}=\omega_{0}+\alpha t \\ &d \theta=\left(\omega_{0}+a t\right) d t \end{aligned} $$ So, if in time \(t\) angular displacenent is \(\theta\) $$ \begin{gathered} \int_{0}^{\theta} d \theta=\int_{0}^{t}\left(\omega_{0}+\alpha t\right) d t \\\ \theta=\omega_{0} t+\frac{1}{2} \alpha t^{2} \end{gathered} $$ Given, \(\alpha=3.0 \mathrm{rads}^{-2}, \omega_{0}=2.0 \mathrm{rads}^{-1}, t=2 \mathrm{~s}\) Hence, or \(\theta=2 \times 2+\frac{1}{2} \times 3 \times(2)^{2}\) \(\theta=4+6=10 \mathrm{rad}\) Eqs. (i) and (ii) are similar to first and second equations of linear motion.So, if in time \(t\) angular displacenent is \(\theta\) $$ \begin{gathered} \int_{0}^{\theta} d \theta=\int_{0}^{t}\left(\omega_{0}+\alpha t\right) d t \\\ \theta=\omega_{0} t+\frac{1}{2} \alpha t^{2} \\ \text { Given, } \alpha=3.0 \mathrm{rads}^{-2}, \omega_{0}=2.0 \mathrm{rads}^{-1}, t=2 \mathrm{~s} \end{gathered} $$ Hence, $$ \theta=2 \times 2+\frac{1}{2} \times 3 \times(2)^{2} $$ or \(\theta=4+6=10 \mathrm{rad}\) Eqs. (i) and (ii) are similar to first and second equations of linear mot?on.

See all solutions

Recommended explanations on Physics 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.