Match list-I with list-II:List-I List-II (A) Kinetic energy of planet \(- \frac{GMm}{a}\) (B) Gravitational Potential energy of Sun-planet system \(- \frac{GMm}{2a}\) (C) Total mechanical energy of planet \(\frac{GM}{r}\) (D) Escape energy at the surface of planet for unit mass object \(- \frac{GMm}{2a}\)
(Where a = radius of planet orbit, r = radius of planet, M = mass of Sun, m = mass of planet) Choose the correct answer from the options given below:
| List-I | List-II |
|---|---|
| (A) Kinetic energy of planet | \(- \frac{GMm}{a}\) |
| (B) Gravitational Potential energy of Sun-planet system | \(- \frac{GMm}{2a}\) |
| (C) Total mechanical energy of planet | \(\frac{GM}{r}\) |
| (D) Escape energy at the surface of planet for unit mass object | \(- \frac{GMm}{2a}\) |
- (A) – II, (B) – I, (C) – IV, (D) – III
- (A) – III, (B) – IV, (C) – I, (D) – II
- (A) – I, (B) – IV, (C) – II, (D) – III
- (A) – I, (B) – II, (C) – III, (D) – IV
The Correct Option is A
Approach Solution - 1
The kinetic energy (KE) of a planet is given by:
\[ \text{KE} = \frac{1}{2} mv^2 = \frac{GMm}{2a} \]
The gravitational potential energy (PE) of the Sun-planet system is:
\[ \text{PE} = -\frac{GMm}{a} \]
The total mechanical energy (TE) of the planet is:
\[ \text{TE} = \text{KE} + \text{PE} = -\frac{GMm}{2a} \]
Escape energy at the surface of the planet for a unit mass object is given by:
\[ \text{Escape Energy} = \frac{Gm}{r} \]
Approach Solution -2
Step 1: Recall the relevant formulas
For a planet of mass \( m \) orbiting the Sun of mass \( M \) at a distance \( a \):
- Kinetic energy of planet, \( K = \dfrac{GMm}{2a} \)
- Gravitational potential energy of the Sun–planet system, \( U = -\dfrac{GMm}{a} \)
- Total mechanical energy of planet, \( E = K + U = -\dfrac{GMm}{2a} \)
- Escape energy per unit mass from surface of planet, \( \varepsilon = \dfrac{GM}{r} \)
Step 2: Match each quantity correctly
| List-I | List-II |
|---|---|
| (A) Kinetic energy of planet | \( +\dfrac{GMm}{2a} \) |
| (B) Gravitational potential energy of Sun–planet system | \( -\dfrac{GMm}{a} \) |
| (C) Total mechanical energy of planet | \( -\dfrac{GMm}{2a} \) |
| (D) Escape energy at the surface of planet for unit mass object | \( +\dfrac{GM}{r} \) |
Step 3: Final correspondence
\[ (A) - II, \quad (B) - I, \quad (C) - IV, \quad (D) - III \]
Final answer
(A) – II, (B) – I, (C) – IV, (D) – III
Top Questions on Gravitation
- Given below are two statements: Statement I: A satellite is moving around earth in an orbit very close to the earth surface. The time period of revolution of satellite depends upon the density of earth.
Statement II: The time period of revolution of the satellite is \[ T = 2\pi \sqrt{\frac{R_e}{g}} \] (for satellite very close to the earth surface), where \(R_e\) is the radius of earth and \(g\) is acceleration due to gravity.
In the light of the above statements, choose the correct answer from the options given below.- JEE Main - 2026
- Physics
- Gravitation
- Initially a satellite of 100 kg is in a circular orbit of radius 1.5\(R_E\). This satellite can be moved to a circular orbit of radius 3\(R_E\) by supplying \(a \times 10^6\) J of energy. The value of a is_________ . (Take Radius of Earth \(R_E = 6 \times 10^6\) m and g = 10 m/s\(^2\)).
- JEE Main - 2026
- Physics
- Gravitation
- The escape velocity from a spherical planet \(A\) is \(10\ \text{km/s}\). The escape velocity from another planet \(B\), whose density and radius are \(10%\) of those of planet \(A\), is _______ m/s.
- JEE Main - 2026
- Physics
- Gravitation
Net gravitational force at the center of a square is found to be \( F_1 \) when four particles having masses \( M, 2M, 3M \) and \( 4M \) are placed at the four corners of the square as shown in figure, and it is \( F_2 \) when the positions of \( 3M \) and \( 4M \) are interchanged. The ratio \( \dfrac{F_1}{F_2} = \dfrac{\alpha}{\sqrt{5}} \). The value of \( \alpha \) is
- JEE Main - 2026
- Physics
- Gravitation
- In the given situation, the force at the center on 1 kg mass is \( F_1 \). Now if \( 4m \) and \( 3m \) are interchanged, the force is \( F_2 \). Given \( \frac{F_1}{F_2} = \frac{2}{\sqrt{\alpha}} \), find \( \alpha \).
- JEE Main - 2026
- Physics
- Gravitation
Questions Asked in JEE Main exam
The equivalent resistance between the points \(A\) and \(B\) in the given circuit is \[ \frac{x}{5}\,\Omega. \] Find the value of \(x\).
- JEE Main - 2026
- Current electricity
Method used for separation of mixture of products (B and C) obtained in the following reaction is:
- JEE Main - 2026
- p -Block Elements
- Consider light travelling from a medium A to medium B separated by a plane interface. If the light undergoes total internal reflection during its travel from medium A to B and the speed of light in media A and B are \(2.4 \times 10^8\) m/s and \(2.7 \times 10^8\) m/s, respectively, then the value of critical angle is :
- JEE Main - 2026
- Newtons Laws of Motion
In the following \(p\text{–}V\) diagram, the equation of state along the curved path is given by \[ (V-2)^2 = 4ap, \] where \(a\) is a constant. The total work done in the closed path is:
- JEE Main - 2026
- Thermodynamics
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.
- JEE Main - 2026
- Coordinate Geometry