The distance of two planets from the sun are $ {10}^{13}$ m and $ {10}^{12}$ m respectively. The ratio of time periods of the planets is
- $\sqrt{10}$
- $ 10 \sqrt{10}$
- 10
- $ \frac{1}{\sqrt{10}}$
The Correct Option is B
Solution and Explanation
Relation between time period (T) and distance of the planet from the sun is $ T^2 \propto r^3 \, \, \,$ or $\, \, \, T \propto r^{3/2}$.
Therefore $\frac{ T_1}{T_2} = \bigg( \frac{r_1}{r_2} \bigg)^{3/2} = \bigg( \frac{{10}^{13}}{{10}^{12}} \bigg)^{3/2} = {10}^{3/2} = 10 \sqrt{10}$.
Top Questions on Gravitation
- At a height \( h \) above the Earth's surface, the acceleration due to gravity becomes \( \frac{g}{\sqrt{3}} \). What is the value of \( h \) in terms of the Earth's radius \( R \)?
- MHT CET - 2025
- Physics
- Gravitation
- The value of \( g \) at height \( h \) above Earth's surface is \( \frac{g}{\sqrt{3}} \). Find \( h \) in terms of the radius of the Earth.
- MHT CET - 2025
- Physics
- Gravitation
- Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R):
(A) A simple pendulum is taken to a planet of mass and radius, 4 times and 2 times, respectively, than the Earth. The time period of the pendulum remains same on earth and the planet.
(R) The mass of the pendulum remains unchanged at Earth and the other planet.
In light of the above statements, choose the correct answer from the options given below:- JEE Main - 2025
- Physics
- Gravitation
A small point of mass \(m\) is placed at a distance \(2R\) from the center \(O\) of a big uniform solid sphere of mass \(M\) and radius \(R\). The gravitational force on \(m\) due to \(M\) is \(F_1\). A spherical part of radius \(R/3\) is removed from the big sphere as shown in the figure, and the gravitational force on \(m\) due to the remaining part of \(M\) is found to be \(F_2\). The value of the ratio \( F_1 : F_2 \) is:
- JEE Main - 2025
- Physics
- Gravitation
- Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R): \begin{itemize} \item[(A)] A simple pendulum is taken to a planet of mass and radius, 4 times and 2 times, respectively, than the Earth. The time period of the pendulum remains same on earth and the planet. \item[(R)] The mass of the pendulum remains unchanged at Earth and the other planet. \end{itemize} In light of the above statements, choose the correct answer from the options given below:
- JEE Main - 2025
- Physics
- Gravitation
Questions Asked in NEET exam
- The E° value for the Mn3+/Mn2+ couple is more positive than that of Cr3+/Cr2+ or Fe3+/Fe2+ due to change of
- NEET (UG) - 2024
- Electronic configuration of atoms and ions
- The compound that will undergo SN1 reaction with the fastest rate is
- NEET (UG) - 2024
- Organic Chemistry- Some Basic Principles and Techniques
- Auxin is used by gardener to prepare weed-free lawns. But no damage is caused to grass as auxin
- NEET (UG) - 2024
- Role of auxin
- In an ecosystem if the Net Primary Productivity (NPP) of first trophic level is 100x (kcal m-2)yr-1, what would be the GPP (Gross Primary Productivity) of the third trophic level of the same ecosystem?
- NEET (UG) - 2024
- Ecosystems
- Arrange the following elements in increasing order of electronegativity:
N, O, F, C, Si
Choose the correct answer from the options given below:- NEET (UG) - 2024
- Modern Periodic Law And The Present Form Of The Periodic Table
Concepts Used:
Gravitation
In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.
Newton’s Law of Gravitation
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
- F ∝ (M1M2) . . . . (1)
- (F ∝ 1/r2) . . . . (2)
On combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2
The dimension formula of G is [M-1L3T-2].