krit.club logo

Gravitation - Universal Law of Gravitation

Grade 9CBSE

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

🔑Concepts

•

The Universal Law of Gravitation states that every object in the universe attracts every other object with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

•

The force acts along the line joining the centers of the two objects.

•

The constant GG is called the Universal Gravitational Constant. Its value was first found by Henry Cavendish using a sensitive torsion balance.

•

The law is 'universal' because it applies to all bodies—whether they are celestial (like stars and planets) or terrestrial (objects on Earth), and whether they are large or small.

•

If the distance dd between two objects is doubled, the gravitational force FF becomes 14\frac{1}{4} of its original value due to the inverse square relationship F∝1d2F \propto \frac{1}{d^2}.

•

Gravitational force is a non-contact force and is responsible for the motion of the Moon around the Earth and the Earth around the Sun.

📐Formulae

F=Gm1m2d2F = G \frac{m_1 m_2}{d^2}

G=6.673×10−11 N m2 kg−2G = 6.673 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}

F∝m1m2F \propto m_1 m_2

F∝1d2F \propto \frac{1}{d^2}

💡Examples

Problem 1:

Calculate the force of gravitation between the Earth and an object of mass 1 kg1 \text{ kg} kept on its surface. Given: Mass of Earth M=6×1024 kgM = 6 \times 10^{24} \text{ kg}, Radius of Earth R=6.4×106 mR = 6.4 \times 10^6 \text{ m}, and G=6.7×10−11 N m2 kg−2G = 6.7 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}.

Solution:

F=6.7×10−11×6×1024×1(6.4×106)2≈9.8 NF = \frac{6.7 \times 10^{-11} \times 6 \times 10^{24} \times 1}{(6.4 \times 10^6)^2} \approx 9.8 \text{ N}

Explanation:

By substituting the values into the formula F=GMmR2F = G \frac{M m}{R^2}, we multiply the masses and the constant GG, then divide by the square of the Earth's radius. The resulting force is approximately 9.8 N9.8 \text{ N}, which represents the weight of a 1 kg1 \text{ kg} mass on Earth.

Problem 2:

What happens to the gravitational force FF between two objects if the mass of one object is doubled and the distance between them is also doubled?

Solution:

Fnew=G(2m1)m2(2d)2=G2m1m24d2=12ForiginalF_{new} = G \frac{(2m_1) m_2}{(2d)^2} = G \frac{2 m_1 m_2}{4 d^2} = \frac{1}{2} F_{original}

Explanation:

Doubling the mass increases the force by a factor of 22, but doubling the distance decreases the force by a factor of 22=42^2 = 4. Therefore, the net effect is 2/4=0.52/4 = 0.5, meaning the force becomes half of its original value.