
Suppose universal gravitational constant starts to decrease, then
(This question may have multiple correct questions)
$(a)$ length of the day on the earth will increase
$(b)$length of the year will increase
$(c)$ the earth will follow a spiral path of decreasing radius
$(d)$ Kinetic energy of the earth will decrease
Answer
180k+ views
Hint: In this question use the direct formula that is $F = G\dfrac{{{m_1}{m_2}}}{{{r^2}}}$, to predict the effect of decrease of G onto the force acting between the earth and the sun. We will observe that on decreasing G the earth will no longer be able to circulate onto a circular orbit around the sun, eventually it becomes spiral. This will help to approach the solution.
Complete step-by-step solution -
As we know that the gravitational force between the two bodies is working as, directly proportional to the product of their individual masses and inversely proportional to the square of the distance between them,
$ \Rightarrow F = G\dfrac{{{m_1}{m_2}}}{{{r^2}}}$N.................... (1)
Where G = universal gravitational constant = 6.674$ \times {10^{ - 11}}$N-m2/Kg2
Now let the first body be earth and the second body be Sun.
So ${m_1}$ is the mass of the earth and ${m_2}$ is the mass of the sun.
And (r) is the distance between them.
Now according to the question if the value of G starts decreasing.
Then from equation (1) the gravitational force between the sun and the earth also starts decreasing.
Therefore earth will follow a spiral path of increasing radius, only then the gravitational force starts decreasing as mass of the sun and the earth remains the same.
So (r) increased, therefore the time taken by the earth to revolve around the sun also increased.
As we know, 1 complete revolution around the sun = 1 year = 365days, when G is constant.
Now when G is decreased, the total time of the revolution around the sun is increased so the length of the year is also increased.
But the rotational motion of the earth on its own axis remains unchanged so the period of the earth’s rotation remains unchanged hence the length of the day on the earth will remain the same.
Now due to decreased G, r will increase so the potential energy (P.E) of the earth also starts increasing.
As the total energy remains the same which is the sum of kinetic energy and potential energy, so if the potential energy of the earth is increased therefore kinetic energy will be decreased so that the total energy remains constant.
Hence option (B) and (D) are the correct answer.
Note – There is often a confusion between g and G. g is the acceleration due to gravity whose value is 9.8 at the surface of the earth however G is the proportionality constant and has a default value of $6.674 \times {10^{11}}mK{g^{ - 1}}{s^{ - 2}}$ It is advised to remember the direct formula for the force of gravitation between two masses that is ${F^1}_g = G\dfrac{{{m_1}{m_2}}}{{{d^2}}}$ as it is very helpful while dealing with forces between two bodies and has involvement of G in it as well.
Complete step-by-step solution -
As we know that the gravitational force between the two bodies is working as, directly proportional to the product of their individual masses and inversely proportional to the square of the distance between them,
$ \Rightarrow F = G\dfrac{{{m_1}{m_2}}}{{{r^2}}}$N.................... (1)
Where G = universal gravitational constant = 6.674$ \times {10^{ - 11}}$N-m2/Kg2
Now let the first body be earth and the second body be Sun.
So ${m_1}$ is the mass of the earth and ${m_2}$ is the mass of the sun.
And (r) is the distance between them.
Now according to the question if the value of G starts decreasing.
Then from equation (1) the gravitational force between the sun and the earth also starts decreasing.
Therefore earth will follow a spiral path of increasing radius, only then the gravitational force starts decreasing as mass of the sun and the earth remains the same.
So (r) increased, therefore the time taken by the earth to revolve around the sun also increased.
As we know, 1 complete revolution around the sun = 1 year = 365days, when G is constant.
Now when G is decreased, the total time of the revolution around the sun is increased so the length of the year is also increased.
But the rotational motion of the earth on its own axis remains unchanged so the period of the earth’s rotation remains unchanged hence the length of the day on the earth will remain the same.
Now due to decreased G, r will increase so the potential energy (P.E) of the earth also starts increasing.
As the total energy remains the same which is the sum of kinetic energy and potential energy, so if the potential energy of the earth is increased therefore kinetic energy will be decreased so that the total energy remains constant.
Hence option (B) and (D) are the correct answer.
Note – There is often a confusion between g and G. g is the acceleration due to gravity whose value is 9.8 at the surface of the earth however G is the proportionality constant and has a default value of $6.674 \times {10^{11}}mK{g^{ - 1}}{s^{ - 2}}$ It is advised to remember the direct formula for the force of gravitation between two masses that is ${F^1}_g = G\dfrac{{{m_1}{m_2}}}{{{d^2}}}$ as it is very helpful while dealing with forces between two bodies and has involvement of G in it as well.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

Trending doubts
Collision - Important Concepts and Tips for JEE

Current Loop as Magnetic Dipole and Its Derivation for JEE

Elastic Collisions in One Dimension - JEE Important Topic

The system shown is just on the verge of slipping The class 11 physics JEE_Main

Find the frictional force between the two blocks in class 11 physics JEE_MAIN

JEE Main Eligibility Criteria 2025

Other Pages
NCERT Solutions For Class 11 Physics Chapter 8 Mechanical Properties Of Solids - 2025-26

JEE Advanced 2025 Revision Notes for Practical Organic Chemistry

Electron Gain Enthalpy and Electron Affinity for JEE

Important Days and Dates in August

MBBS Seats in India 2025: State & College Wise Updates

NEET Cut Off Tamil Nadu 2025: Govt & Private MBBS/BDS College List
