Question:
The escape velocity of a body from the earth depends on
(i) the mass of the body.
(ii) the location from where it is projected.
(iii) the direction of projection.
(iv) the height of the location from where the body is launched.
The escape velocity of a body from the earth depends on
(i) the mass of the body.
(ii) the location from where it is projected.
(iii) the direction of projection.
(iv) the height of the location from where the body is launched.
Updated On: Jul 7, 2022
- (i) and (ii)
- (ii) and (iv)
- (i) and (iii)
- (iii) and (iv)
Hide Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
The escape velocity is independent of mass of the body and the direction of projection. It depends upon the gravitational potential at the point from where the body is launched. Since this potential depends slightly on the latitude and height of the point, the escape velocity depends slightly on these factors.
Was this answer helpful?
0
0
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
View More Questions
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].