Separation between earth and sun is given by 1.5 × 106 km. Time period of another planet is 2.83 year. Find distance of another planet from sun?
Separation between earth and sun is given by 1.5 × 106 km. Time period of another planet is 2.83 year. Find distance of another planet from sun?
3 × 106 km
2 × 107 km
3 × 107 km
2 × 106 km
The Correct Option is A
Solution and Explanation
T2 ∝ R3
(T1/T2)2 = (R1/R2)3
\((\frac{1}{2.83})^2\) = {(1.5 x 106)/R2}3
R2 = \((1.5 × 10^{6}) (2.83)\frac{2}{3}\) km
= \((1.5 × 10^{6}) (8)\frac{1}{3}\)
= 3 × 106 km
The correct option is (A): 3 × 106 km
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
- Two objects of masses 5 kg and 10 kg are placed 2 meters apart. What is the gravitational force between them?
(Use \(G = 6.67 \times 10^{-11}\, \mathrm{Nm^2/kg^2}\))- AP EAPCET - 2025
- Physics
- Gravitation
- The escape velocity from the surface of a planet is \(v_e\). What will be the escape velocity from a planet whose mass and radius are twice that of the original planet?
- BITSAT - 2025
- Physics
- Gravitation
- Consider a star of mass $ m_2 $ kg revolving in a circular orbit around another star of mass $ m_1 $ kg with $ m_1 \gg m_2 $. The heavier star slowly acquires mass from the lighter star at a constant rate of $ \gamma $ kg/s. In this transfer process, there is no other loss of mass. If the separation between the centers of the stars is $ r $, then its relative rate of change $ \frac{1}{r} \frac{dr}{dt} $ (in s$^{-1}$) is given by:
- JEE Advanced - 2025
- Physics
- Gravitation
- If the weight of an object is 200 Newtons, then the weight of the object at the midpoint between the Earth's surface and the Earth's center will be:
- UPCATET - 2025
- Physics
- Gravitation
Questions Asked in JEE Main exam
- A force of 49 N acts tangentially at the highest point of a sphere (solid of mass 20 kg) kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is:
- JEE Main - 2025
- Quantum Mechanics
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
- JEE Main - 2025
- Relations and functions
- If \( \alpha + i\beta \) and \( \gamma + i\delta \) are the roots of the equation \( x^2 - (3-2i)x - (2i-2) = 0 \), \( i = \sqrt{-1} \), then \( \alpha\gamma + \beta\delta \) is equal to:
- JEE Main - 2025
- Complex numbers
- The sum of all local minimum values of the function \( f(x) \) as defined below is:
\[ f(x) = \begin{cases} 1 - 2x & \text{if } x < -1, \\[10pt] \frac{1}{3}(7 + 2|x|) & \text{if } -1 \leq x \leq 2, \\[10pt] \frac{11}{18}(x-4)(x-5) & \text{if } x > 2. \end{cases} \]- JEE Main - 2025
- Functions
- Let \( \langle a_n \rangle \) be a sequence such that \( a_0 = 0 \), \( a_1 = \frac{1}{2} \), and \( 2a_{n+2} = 5a_{n+1} - 3a_n \).n= 0,1,2,3.... Then \( \sum_{k=1}^{100} a_k \) is equal to:
- JEE Main - 2025
- Sequences and Series
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].