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Newton's Laws of Motion - Gravitation-advanced

Grade 9CBSE

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

🔑Concepts

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The Universal Law of Gravitation states that every object in the universe attracts every other object 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 (dd) between them: F∝m1m2d2F \propto \frac{m_1 m_2}{d^2}.

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The Universal Gravitational Constant (GG) is a scalar quantity with a value of 6.673×10−11 N m2 kg−26.673 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}. Its value remains constant throughout the universe.

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Acceleration due to gravity (gg) is the acceleration gained by an object due to the gravitational force of a celestial body. On Earth, g=GMR2g = \frac{GM}{R^2}, where MM is the mass of Earth and RR is its radius.

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Variation of gg: The value of gg is not constant everywhere on Earth. It is greater at the poles and lesser at the equator because the Earth's radius is smaller at the poles (Rp<ReR_p < R_e).

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Mass is the measure of inertia and remains constant everywhere in the universe. Weight (WW) is the force with which an object is attracted towards the center of the Earth, given by W=mgW = mg.

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Kepler's Third Law (Law of Periods): The square of the time period of revolution of a planet (TT) is directly proportional to the cube of the semi-major axis of its orbit (rr), expressed as T2∝r3T^2 \propto r^3.

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Free Fall: When an object falls towards Earth under the sole influence of gravity, it is said to be in free fall. The equations of motion are modified by replacing acceleration (aa) with gg.

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The value of gg decreases with altitude (height above Earth's surface) and with depth (towards the center of the Earth). At the center of the Earth, g=0g = 0.

📐Formulae

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

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

W=m×gW = m \times g

v=u+gtv = u + gt

s=ut+12gt2s = ut + \frac{1}{2}gt^2

v2=u2+2gsv^2 = u^2 + 2gs

T2r3=constant\frac{T^2}{r^3} = \text{constant}

gh=g(1+hR)−2≈g(1−2hR) for h≪Rg_h = g \left(1 + \frac{h}{R}\right)^{-2} \approx g \left(1 - \frac{2h}{R}\right) \text{ for } h \ll R

💡Examples

Problem 1:

Calculate the gravitational force between the Earth and an object of mass 1 kg1 \text{ kg} on its surface. (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}, G=6.7×10−11 N m2 kg−2G = 6.7 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2})

Solution:

Given: m1=6×1024 kgm_1 = 6 \times 10^{24} \text{ kg}, m2=1 kgm_2 = 1 \text{ kg}, d=6.4×106 md = 6.4 \times 10^6 \text{ m}. Using the formula: F=Gm1m2d2F = G \frac{m_1 m_2}{d^2} F=6.7×10−11×6×1024×1(6.4×106)2F = \frac{6.7 \times 10^{-11} \times 6 \times 10^{24} \times 1}{(6.4 \times 10^6)^2} F=40.2×101340.96×1012F = \frac{40.2 \times 10^{13}}{40.96 \times 10^{12}} F≈9.81 NF \approx 9.81 \text{ N}

Explanation:

This force is equal to the weight of the 1 kg1 \text{ kg} object on the surface of the Earth, which confirms that g≈9.8 m/s2g \approx 9.8 \text{ m/s}^2.

Problem 2:

If a planet has a mass double that of Earth and a radius three times that of Earth, what would be the acceleration due to gravity on that planet compared to Earth's gg?

Solution:

Let MM and RR be the mass and radius of Earth. ge=GMR2g_e = \frac{GM}{R^2}. For the new planet: M′=2MM' = 2M and R′=3RR' = 3R. g′=G(2M)(3R)2g' = \frac{G(2M)}{(3R)^2} g′=2GM9R2g' = \frac{2GM}{9R^2} g′=29geg' = \frac{2}{9} g_e

Explanation:

Since gravity is directly proportional to mass and inversely proportional to the square of the radius, doubling the mass increases gravity by 2×2\times, but tripling the radius decreases gravity by 32=9×3^2 = 9\times.

Problem 3:

A ball is thrown vertically upwards and rises to a height of 20 m20 \text{ m}. Calculate the velocity with which the object was thrown upwards. (Take g=10 m/s2g = 10 \text{ m/s}^2)

Solution:

Given: s=20 ms = 20 \text{ m}, v=0 m/sv = 0 \text{ m/s} (at highest point), g=−10 m/s2g = -10 \text{ m/s}^2 (upward motion). Using v2=u2+2gsv^2 = u^2 + 2gs: 0=u2+2(−10)(20)0 = u^2 + 2(-10)(20) 0=u2−4000 = u^2 - 400 u2=400u^2 = 400 u=20 m/su = 20 \text{ m/s}

Explanation:

The negative sign for gg is used because the gravitational force acts in the opposite direction to the displacement of the ball.