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Physics · Chapter

Gravitation

This chapter introduces the fundamental force of gravitation, beginning with Newton's Law of Universal Gravitation. You will learn to calculate the force of attraction between any two objects with mass. The chapter then develops the concepts of gravitational field, potential, and potential energy, which are essential tools for analyzing the effects of gravity without constantly referring back to forces.

These principles are applied to large-scale systems, primarily the Earth and other celestial bodies. A significant part of the chapter deals with the acceleration due to gravity, 'g', and how it varies with altitude, depth, and the rotation of the Earth. You will study the motion of satellites, calculating their orbital velocity, time period, and total energy. The chapter also covers escape velocity, the minimum speed required for an object to break free from a planet's gravitational pull.

The study concludes with Kepler's three laws of planetary motion, which describe the orbits of planets around the sun. You will see how these empirical laws are a direct consequence of Newton's law of gravitation. Students often find the problems involving the variation of 'g' and the energy calculations for satellites challenging. The chapter relies heavily on concepts from Work, Energy, and Power, as well as Circular Motion.

Formulas for Gravitation

Newton's Law of Universal Gravitation
F=Gm1m2r2F = \frac{Gm_1m_2}{r^2} G is the universal gravitational constant, r is the distance between the centers of masses m1 and m2.
Acceleration due to gravity on a planet's surface
g=GMR2g = \frac{GM}{R^2} M is the mass of the planet, R is its radius.
Variation of g with altitude h
gh=g(1+hR)−2g_h = g\left(1 + \frac{h}{R}\right)^{-2} For h << R, this can be approximated as g_h \approx g(1 - 2h/R).
Variation of g with depth d
gd=g(1−dR)g_d = g\left(1 - \frac{d}{R}\right) This assumes the planet has a uniform density.
Gravitational Potential Energy
U=−GMmrU = -\frac{GMm}{r} The zero of potential energy is taken at infinite separation (r = ∞).
Gravitational Potential
V=−GMrV = -\frac{GM}{r} V is the work done to bring a unit mass from infinity to a point.
Escape Velocity
ve=2GMR=2gRv_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR} This is the minimum speed to escape a planet's gravity from its surface.
Orbital Velocity of a satellite
vo=GMrv_o = \sqrt{\frac{GM}{r}} r is the radius of the circular orbit from the planet's center (r = R+h).
Time Period of a satellite
T2=4π2GMr3T^2 = \frac{4\pi^2}{GM}r^3 This is a form of Kepler's Third Law for circular orbits.
Total Energy of an orbiting satellite
E=−GMm2rE = -\frac{GMm}{2r} For a satellite in a stable orbit, Total Energy = - (Kinetic Energy).
Kepler's Third Law
T2∝a3T^2 \propto a^3 'a' is the semi-major axis of the elliptical orbit.

One of these questions, worked through

JEE Advanced 2015

A large spherical mass M is fixed at one position and two identical point masses m are kept on a line passing through the centre of M (see figure). The point masses are connected by a rigid massless rod of length ℓ and this assembly is free to move along the line connecting them. All three masses interact only through their mutual gravitational interaction. When the point mass nearer to M is at a distance r=3ℓr = 3ℓ from M, the tension in the rod is zero for m=k(M/288).m = k(M/288). The value of k is ______.
Watch this solved on the board The full walkthrough, written out a line at a time with narration explaining each step as it appears. Pause anywhere and ask why a line is there. Free to watch

Or read the worked solution below. Jump to the steps ↓

Step 1: Identify the positions of the masses.

The nearer point mass is at a distance of 3ℓ3\ell from MM, and the farther point mass is at a distance of 4ℓ4\ell from MM.

Step 2: Set up the condition for zero tension in the rod.

For the tension in the rod to be zero, the accelerations of both point masses along the line must be equal.

a1=a2a_1 = a_2

Step 3: Write the acceleration of each point mass.

Taking the direction towards MM as negative, write the gravitational acceleration of each mass.

a1=−GM(3ℓ)2+Gmℓ2a_1 = -\frac{GM}{(3\ell)^2} + \frac{Gm}{\ell^2} a2=−GM(4ℓ)2−Gmℓ2a_2 = -\frac{GM}{(4\ell)^2} - \frac{Gm}{\ell^2}

Step 4: Equate the accelerations and solve for mm.

Set a1=a2a_1 = a_2 to find the relation between mm and MM.

−GM9ℓ2+Gmℓ2=−GM16ℓ2−Gmℓ2-\frac{GM}{9\ell^2} + \frac{Gm}{\ell^2} = -\frac{GM}{16\ell^2} - \frac{Gm}{\ell^2} 2Gmℓ2=GMℓ2(19−116)\frac{2Gm}{\ell^2} = \frac{GM}{\ell^2}\left(\frac{1}{9} - \frac{1}{16}\right) 2m=M(7144)2m = M\left(\frac{7}{144}\right) m=7M288m = \frac{7M}{288}

Step 5: Determine the value of kk.

Compare the expression for mm with the given relation m=k(M288)m = k\left(\frac{M}{288}\right).

k=7k = 7

Answer: 7

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