Which of the following curves represents the variation of total energy with radius r for a satellite in a circular orbit?
- P
- Q
- R
- S
The Correct Option is C
Solution and Explanation
Top Questions on 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
- "The height from Earth's surface at which acceleration due to gravity becomes \( g/4 \), where \( g \) is acceleration due to gravity on the surface of Earth and \( R \) is the radius of Earth?"
- MHT CET - 2024
- Physics
- Gravitation
The height from Earth's surface at which acceleration due to gravity becomes \(\frac{g}{4}\) is \(\_\_\)? (Where \(g\) is the acceleration due to gravity on the surface of the Earth and \(R\) is the radius of the Earth.)
- MHT CET - 2024
- Physics
- Gravitation
Questions Asked in AMUEEE exam
- The ratio of the half-life time $ (t_{1/2}) $ , to the three quarter life-time, $ (t_{3/4}) $ , for a reaction that is second order
- AMUEEE - 2018
- Order of Reaction
- $ 0.002\, M $ solution of a weak acid has an equivalent conductance $ (\wedge) 60\, ohm^{-1}\, cm^{2}\, eq^{-1} $ . What will be the $ pH $ ? (Given : $ \wedge = 400 \, ohm^{-1} \, cm^{2} \, eq^{-1} $ ]
- AMUEEE - 2018
- Electrochemistry
- The electrode potential, $ E^{\circ} $ , for the reduction of $ MnO^{-}_{4} $ to $ Mn^{2+} $ in acidic medium is $ +1.51\, V $ . Which of the following metal(s) will be oxidised? The reduction reactions and standard electrode potentials for $ Zn^{2+}, Ag^{+} $ , and $ Au^{+} $ are given as $ Zn^{2+} (aq) +2e \to Zn(s), E^{\circ} = -0.762\, V $ $ Ag+ (aq) +e {\rightleftharpoons} Ag (s), E^{\circ}=+0.80 \,V $ $ Au^{+} (aq) +e {\rightleftharpoons} Au(s), E^{\circ}+1.69\,V $
- AMUEEE - 2018
- Electrochemistry
- The $ EAN $ value $ y \left[Ti\left(\sigma-C_{6}H_{5}\right)_{2}\left(\pi-C_{5}H_{5}\right)_{2}\right]^{0} $ is
- AMUEEE - 2018
- coordination compounds
- The only force acting on a $ 2 \,kg $ body that is moving in $ xy- $ plane has a magnitude of $ 5\, N $ . The body initially has a velocity of $ 4 \,m/s $ in the positive $ x $ -direction. Some time later, the body has a velocity of $ 6 \,m/s $ in the positive $ y- $ direction. The work done on the body by the $ 5 \,N $ force during this time is
- AMUEEE - 2018
- work, energy and power
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].