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Gravitation - Universal Law of Gravitation

Grade 11ICSEPhysics

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

🔑Concepts

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Newton's Universal Law of Gravitation states that every particle in the universe attracts every other particle with a force which is directly proportional to the product of their masses (m1m2m_1 m_2) and inversely proportional to the square of the distance (rr) between their centers.

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The Gravitational Constant (GG) is a universal constant. Its value is approximately 6.674×10−11 N m2 kg−26.674 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2} and its dimensional formula is [M−1L3T−2][M^{-1} L^3 T^{-2}].

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The gravitational force is a central force, meaning it acts along the line joining the centers of the two interacting bodies.

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The force is always attractive in nature and independent of the medium between the two masses.

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Principle of Superposition: The total gravitational force exerted on a point mass by a system of masses is the vector sum of the gravitational forces exerted by each individual mass.

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Gravitational force follows the inverse square law, where F∝1r2F \propto \frac{1}{r^2}.

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Acceleration due to gravity (gg) on the surface of a planet of mass MM and radius RR is given by g=GMR2g = \frac{GM}{R^2}.

📐Formulae

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

F⃗12=−Gm1m2r2r^21\vec{F}_{12} = -\frac{G m_1 m_2}{r^2} \hat{r}_{21}

G=6.67×10−11 N m2 kg−2G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}

g=GMR2g = \frac{GM}{R^2}

F1F2=(r2r1)2 (for constant masses)\frac{F_1}{F_2} = \left(\frac{r_2}{r_1}\right)^2 \text{ (for constant masses)}

💡Examples

Problem 1:

Calculate the gravitational force of attraction between two metal spheres each of mass 50 kg50 \text{ kg} if the distance between their centers is 50 cm50 \text{ cm}.

Solution:

Given: m1=50 kgm_1 = 50 \text{ kg}, m2=50 kgm_2 = 50 \text{ kg}, r=50 cm=0.5 mr = 50 \text{ cm} = 0.5 \text{ m}, G=6.67×10−11 N m2 kg−2G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}. Using F=Gm1m2r2F = G \frac{m_1 m_2}{r^2}: F=6.67×10−11×50×50(0.5)2F = 6.67 \times 10^{-11} \times \frac{50 \times 50}{(0.5)^2} F=6.67×10−11×25000.25F = 6.67 \times 10^{-11} \times \frac{2500}{0.25} F=6.67×10−11×10000F = 6.67 \times 10^{-11} \times 10000 F=6.67×10−7 NF = 6.67 \times 10^{-7} \text{ N}

Explanation:

Substitute the values into the Universal Law of Gravitation formula. Ensure all units are in SI (convert cm to m) before calculation.

Problem 2:

A planet has a mass twice that of Earth and a radius three times that of Earth. Find the acceleration due to gravity on this planet if gg on Earth is 9.8 m/s29.8 \text{ m/s}^2.

Solution:

Let MeM_e and ReR_e be mass and radius of Earth. Mp=2MeM_p = 2M_e and Rp=3ReR_p = 3R_e. Acceleration due to gravity is g=GMR2g = \frac{GM}{R^2}. So, gp=G(2Me)(3Re)2=29(GMeRe2)g_p = \frac{G(2M_e)}{(3R_e)^2} = \frac{2}{9} \left( \frac{GM_e}{R_e^2} \right). Since GMeRe2=ge\frac{GM_e}{R_e^2} = g_e, then gp=29×9.8≈2.18 m/s2g_p = \frac{2}{9} \times 9.8 \approx 2.18 \text{ m/s}^2.

Explanation:

The acceleration due to gravity is directly proportional to the mass of the planet and inversely proportional to the square of its radius.