krit.club logo

Gravitation - Universal Law of Gravitation

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Newton's Universal Law of Gravitation states that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses (m1m_1 and m2m_2) and inversely proportional to the square of the distance (rr) between them.

•

The force acts along the line joining the centers of the two particles, making it a central force.

•

The Universal Gravitational Constant GG is independent of the nature of the masses, the medium between them, and time. Its value is 6.674×10−11 N m2 kg−26.674 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}.

•

Gravitational force is a conservative force, meaning the work done by it in moving a particle between two points is independent of the path taken.

•

The law follows the Principle of Superposition: the total gravitational force on a particle due to a number of other particles is the vector sum of the forces exerted by the individual particles: F⃗net=∑F⃗i\vec{F}_{net} = \sum \vec{F}_i.

•

Gravitational force is always attractive and forms an action-reaction pair satisfying Newton's third law: F⃗12=−F⃗21\vec{F}_{12} = -\vec{F}_{21}.

📐Formulae

F=Gm1m2r2F = G \frac{m_1 m_2}{r^2}

F⃗12=−Gm1m2r2r^12\vec{F}_{12} = -G \frac{m_1 m_2}{r^2} \hat{r}_{12}

G=Fr2m1m2G = \frac{F r^2}{m_1 m_2}

[G]=[M−1L3T−2][G] = [M^{-1} L^3 T^{-2}]

💡Examples

Problem 1:

Calculate the gravitational force of attraction between two spheres of mass 20 kg20 \text{ kg} and 40 kg40 \text{ kg} respectively, if the distance between their centers is 2 m2 \text{ m}. (Take G=6.67×10−11 N m2 kg−2G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2})

Solution:

F=6.67×10−11×20×4022=6.67×10−11×8004=6.67×10−11×200=1.334×10−8 NF = 6.67 \times 10^{-11} \times \frac{20 \times 40}{2^2} = 6.67 \times 10^{-11} \times \frac{800}{4} = 6.67 \times 10^{-11} \times 200 = 1.334 \times 10^{-8} \text{ N}.

Explanation:

The Universal Law of Gravitation formula F=Gm1m2r2F = G \frac{m_1 m_2}{r^2} is used by substituting the given masses m1=20 kgm_1 = 20 \text{ kg}, m2=40 kgm_2 = 40 \text{ kg}, and distance r=2 mr = 2 \text{ m}.

Problem 2:

If the distance between two masses is doubled, what happens to the gravitational force between them?

Solution:

The new force F′=Gm1m2(2r)2=14(Gm1m2r2)=F4F' = G \frac{m_1 m_2}{(2r)^2} = \frac{1}{4} \left( G \frac{m_1 m_2}{r^2} \right) = \frac{F}{4}.

Explanation:

Since the force is inversely proportional to the square of the distance (F∝1r2F \propto \frac{1}{r^2}), doubling the distance (r→2rr \to 2r) results in the force becoming one-fourth of its original value.

Universal Law of Gravitation - Revision Notes & Key Formulas | CBSE Class 11 Physics