Suppose a situation in which two planets orbits around the sun in the same orbit. If the mass of plant 1 is twice the mass of planet 2, then what do they have same?
- Potential energy
- Kinetic energy
- Total energy
- Velocity
The Correct Option is A
Solution and Explanation
The correct option is (D): Velocity
Top Questions on Gravitation
- Three identical spheres of mass m, are placed at the vertices of an equilateral triangle of length a. When released, they interact only through gravitational force and collide after a time T = 4 seconds. If the sides of the triangle are increased to length 2a and also the masses of the spheres are made 2m, then they will collide after ______ seconds.
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- A particle is released from height S above the surface of the earth. At certain height its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively.
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Match the LIST-I with LIST-II
\[ \begin{array}{|l|l|} \hline \text{LIST-I} & \text{LIST-II} \\ \hline \text{A. Gravitational constant} & \text{I. } [LT^{-2}] \\ \hline \text{B. Gravitational potential energy} & \text{II. } [L^2T^{-2}] \\ \hline \text{C. Gravitational potential} & \text{III. } [ML^2T^{-2}] \\ \hline \text{D. Acceleration due to gravity} & \text{IV. } [M^{-1}L^3T^{-2}] \\ \hline \end{array} \]
Choose the correct answer from the options given below:
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- A satellite is orbiting the Earth at a height of \( 10^4 \, \text{km} \) above the Earth's surface. If the radius of the Earth is \( 6.4 \times 10^6 \, \text{m} \), calculate the orbital speed of the satellite. (Gravitational constant \( G = 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2 \) and Earth's mass \( M = 6 \times 10^{24} \, \text{kg} \))
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- A body of mass $ 1.5 \, \text{kg} $ is dropped from a height of $ 20 \, \text{m} $. What is its speed just before hitting the ground? (Assume $ g = 9.8 \, \text{m/s}^2 $)
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Questions Asked in JEE Main exam
In the following circuit, the reading of the ammeter will be: (Take Zener breakdown voltage = 4 V)
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If $10 \sin^4 \theta + 15 \cos^4 \theta = 6$, then the value of $\frac{27 \csc^6 \theta + 8 \sec^6 \theta}{16 \sec^8 \theta}$ is:
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- Trigonometric Identities
If the area of the region $\{ (x, y) : |x - 5| \leq y \leq 4\sqrt{x} \}$ is $A$, then $3A$ is equal to
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Let $A = \begin{bmatrix} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{bmatrix}$. If for some $\theta \in (0, \pi)$, $A^2 = A^T$, then the sum of the diagonal elements of the matrix $(A + I)^3 + (A - I)^3 - 6A$ is equal to
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Let $A = \{ z \in \mathbb{C} : |z - 2 - i| = 3 \}$, $B = \{ z \in \mathbb{C} : \text{Re}(z - iz) = 2 \}$, and $S = A \cap B$. Then $\sum_{z \in S} |z|^2$ is equal to
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- Determinants
Concepts Used:
Keplers Laws
Kepler’s laws of planetary motion are three laws describing the motion of planets around the sun.
Kepler First law – The Law of Orbits
All the planets revolve around the sun in elliptical orbits having the sun at one of the foci.
Kepler’s Second Law – The Law of Equal Areas
It states that the radius vector drawn from the sun to the planet sweeps out equal areas in equal intervals of time.
Kepler’s Third Law – The Law of Periods
It states that the square of the time period of revolution of a planet is directly proportional to the cube of its semi-major axis.
T2 ∝ a3