JEE Physics Gravitation Notes Set 05

Gravitational Force and Superposition

Two point masses attract each other along the line joining them with \(F = \frac{Gm_1m_2}{r^2}\), \(G = 6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2}\). The force is central, conservative and does not depend on the medium between the masses. With several masses, the force on one is the vector sum of the forces from each of the others (superposition). A uniform solid sphere or thin shell attracts an outside point as if all its mass were at the centre.

Worked example: for two equal masses \(m\) at the ends of a line of length \(l\), the field is zero at the midpoint but the potential there is \(-\frac{4Gm}{l}\). Fields cancel as vectors; potentials add as numbers.

Gravitational Field and Potential

1. Field (intensity) at a point is the force per unit mass: \(\vec E = -\frac{GM}{r^2}\hat r\) for a point mass.
2. Potential is the work done by an external agent per unit mass in bringing a unit mass from infinity: \(V = -\frac{GM}{r}\). For several masses, \(V = \sum -\frac{Gm_i}{r_i}\).
3. Relation: \(\vec E = -\nabla V\), or \(E = -\frac{dV}{dr}\). The field points down the potential and is perpendicular to equipotential surfaces. If \(V = 20x + 40y\) then \(\vec E = -20\hat i - 40\hat j\).
4. If \(E = -\frac{k}{r}\), then \(V = k\ln\frac{r}{d_i} + V_i\), where the potential is \(V_i\) at \(r = d_i\).

Field and Potential of Standard Objects

1. The field of a ring on its axis is zero at the centre and at infinity, with a maximum \(\frac{2GM}{3\sqrt3\,a^2}\) at \(x = \pm\frac{a}{\sqrt2}\).
2. A thin shell makes the field zero inside, but the potential inside is not zero; it is equal to the value on the surface.
3. For a solid sphere the field increases linearly from the centre to the surface and then falls as \(\frac{1}{r^2}\). The field is greatest at the surface.
4. A cavity inside a uniform sphere is handled by superposition: add a sphere of negative mass in the cavity. Inside a spherical cavity the field is uniform and equal to \(\frac{4}{3}\pi G\rho\,\vec d\), where \(\vec d\) joins the centre of the sphere to the centre of the cavity.
5. If the density varies as \(\frac{1}{r}\), the mass inside radius \(r\) is proportional to \(r^2\), so the field is the same at all points inside.

Worked example: a solid lead sphere of mass \(M\) and radius \(R\) has a spherical cavity of radius \(\frac{R}{2}\) whose surface passes through the centre. A small mass at distance \(2R\) from the centre on the line through the cavity feels \(\frac{GMm}{4R^2} - \frac{G(M/8)m}{(3R/2)^2} = \frac{7GMm}{36R^2}\).

Potential Energy and Self Energy

1. Potential energy of two masses: \(U = -\frac{Gm_1m_2}{r}\). Change in energy between two distances is the work done by an external agent: \(\Delta U = GMm\left(\frac{1}{r_1} - \frac{1}{r_2}\right)\).
2. Near the earth, \(\Delta U = mgh\) only for \(h \ll R\). In general \(\Delta U = \frac{mgh}{1 + h/R}\). Lifting a body to a height \(R\) gives \(\frac{mgR}{2}\), and to a height \(3R\) gives \(\frac{3mgR}{4}\).
3. Self energy of a system of particles is the sum over pairs: \(U = -\sum_{pairs}\frac{Gm_im_j}{r_{ij}}\). Three equal masses at the corners of a triangle of side \(a\) have \(U = -\frac{3Gm^2}{a}\).
4. Self energy of a uniform sphere: \(U = -\frac{3GM^2}{5R}\); this is the energy needed to disperse all its parts to infinity. For the earth it is about \(2\times10^{32}\) J.
5. Self energy of a thin uniform shell: \(-\frac{GM^2}{2R}\).

Variation of g

If the earth stopped rotating, \(g\) at the equator would increase by \(\omega^2R\), about 0.034 m/s². If the earth rotated so fast that \(g_{eq} = 0\), the day would be about 84 minutes long. A pendulum is a little slower at the equator than at the poles for this reason.

Escape Speed

1. \(v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR} = \sqrt{\frac{8\pi G\rho}{3}}\,R\). For the earth it is 11.2 km/s, for the moon 2.4 km/s.
2. It depends on the mass and radius of the planet (and the starting distance) but not on the mass of the body or the direction of projection (if the path does not hit the planet).
3. Escape speed is \(\sqrt2\) times the circular orbital speed at the same distance. A satellite in a circular orbit needs its kinetic energy doubled to escape.
4. Projected radially at \(kv_e\) (\(k < 1\)): the greatest distance from the centre is \(\frac{R}{1 - k^2}\).
5. Planets with the same density have escape speed proportional to their radius.

Kepler's Laws

1. Law of orbits: planets move in ellipses with the sun at a focus.
2. Law of areas: the areal velocity \(\frac{dA}{dt} = \frac{L}{2m}\) is constant, because the force is central and angular momentum is conserved. A planet is fastest at the closest point (perihelion) and slowest at the farthest (aphelion), with \(v_1r_1 = v_2r_2\) there.
3. Law of periods: \(T^2 = \frac{4\pi^2}{GM}a^3\), where \(a\) is the semi-major axis. A graph of \(\log T\) against \(\log a\) is a straight line of slope \(\frac{3}{2}\).
4. At the ends of the major axis, \(v_{max}^2 = \frac{2GM\,r_{far}}{r_{near}(r_{near}+r_{far})}\), and the total energy of the orbit is \(-\frac{GMm}{2a}\).

Satellites

1. Circular orbit of radius \(r\): \(v = \sqrt{\frac{GM}{r}}\), \(T = 2\pi\sqrt{\frac{r^3}{GM}}\). Near the earth \(v \approx 7.9\) km/s and \(T \approx 84\) minutes.
2. Energies: \(KE = \frac{GMm}{2r}\), \(PE = -\frac{GMm}{r}\), \(E = -\frac{GMm}{2r} = -KE\). A satellite that loses energy to air drag moves to a lower orbit and speeds up.
3. Binding energy of a satellite is \(\frac{GMm}{2r}\). Energy needed to put a satellite in an orbit of radius \(r\) from the surface at rest: \(\frac{GMm}{R} - \frac{GMm}{2r}\). For \(r = 3R\) this is \(\frac{5GMm}{6R}\).
4. Everything inside a satellite is in free fall, so it appears weightless, though gravity still acts on it.
5. A package released from a satellite stays in the same orbit and moves along with it.

Geostationary Satellite

A geostationary satellite stays above one point on the equator. It must orbit in the equatorial plane from west to east with a period of 24 h (about 23 h 56 min with respect to the stars). Its radius from the earth's centre is about 42000 km (height about 36000 km, or about \(6R\)), its orbital speed about 3.1 km/s, and all such satellites have the same radius and speed. If the earth's angular speed doubled, the radius would fall by the factor \(4^{1/3}\).

Path Depends on the Speed of Projection

A body at distance \(r\) from the centre of the earth with a speed \(v\) perpendicular to the radius, where \(v_0 = \sqrt{\frac{GM}{r}}\):

Binary Stars

Two stars of masses \(M_1\), \(M_2\) separated by \(d\) orbit their common centre of mass with the same angular speed. Their orbit radii satisfy \(M_1R_1 = M_2R_2\) (the lighter star has the larger orbit), and the period is \(T = 2\pi\sqrt{\frac{d^3}{G(M_1+M_2)}}\). For masses \(m\) and \(2m\), the ratio of their angular momenta equals the ratio of their kinetic energies, both equal to 2 (heavier : lighter = 1 : 2).

Common Mistakes

1. Using height instead of distance from the centre in the inverse square law.
2. Using \(mgh\) for large heights.
3. Forgetting that field is zero inside a shell but potential is not.
4. Using the separation of two stars as the radius of their orbits.