Question:
Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?
Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?
Updated On: Jan 9, 2024
Hide Solution
Verified By Collegedunia
Solution and Explanation
All objects fall on ground with constant acceleration, called acceleration due to gravity (in the absence of air resistances). It is constant and does not depend upon the mass of an object. Hence, heavy objects do not fall faster than light objects.
Was this answer helpful?
1
0
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
View More Questions