Chapter 13 Gravitation
e = c/a A note about eccentricity For a circle c = 0 à e = 0 a
Orbit Examples Mercury has the highest eccentricity of any planet (a) e Mercury = 0.21 Halley s comet has an orbit with high eccentricity (b) e Halley s comet = 0.97
Notes About Ellipses, Planet Orbits The Sun is at one focus Nothing is located at the other focus Aphelion is the point farthest away from the Sun The distance for aphelion is a + c For an orbit around the Earth, this point is called the apogee Perihelion is the point nearest the Sun The distance for perihelion is a c For an orbit around the Earth, this point is called the perigee a
Kepler s First Law A circular orbit is a special case of the general elliptical orbits Elliptical (and circular) orbits are allowed for bound objects A bound object repeatedly orbits the center An unbound object would pass by and not return These objects could have paths that are parabolas (e = 1) and hyperbolas (e > 1)
Kepler s Second Law Is a consequence of conservation of angular momentum The force produces no torque, so angular momentum is conserved! L =! r x! p = M P! r x! v = constant
Kepler s Second Law, cont. Geometrically, in a time dt, the radius vector r sweeps out the area da, which is half the area of the parallelogram 1 2! r x d r! Its displacement is given by d! r =! vdt
Kepler s Second Law, final Mathematically, we can say da dt L = = constant The law applies to any central force, whether inverse-square or not Area = 1 2! r x d! r d! r =! vdt Way 2 : A = ½ r x dr = ½ r x vdt 2M p Way 1 : da = ½ r x dr = ½ r x vdt da/dt = ½ (r x v) + ½ ( dr/dt x v dt) Prove this yourself using = ½ this (r x v) + ½ (v x v dt) da/dt = ½ (r x v) [ as v x v =0 ] = ½ ( r x v) M p /M p = ½ ( r x v) M p /M p = L/2M p = L/2M p
Newton s Law of Universal Gravitation Every particle in the Universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the distance between them mm 1 2 Fg = G r 2 G is the universal gravitational constant and equals 6.673 x 10-11 Nm 2 / kg 2
Kepler s Third Law T 3 /a 3 = constant Can be predicted from the inverse square law Start by assuming a circular orbit The gravitational force supplies a centripetal force K s is a constant GM Sun M Planet = M v 2 Planet r 2 r v = 2πr T! T 2 4π 2 $ = # & r 3 = K S r 3 " % GM Sun
Kepler s Third Law, cont This can be extended to an elliptical orbit Replace r with a Remember a is the semimajor axis T " 4π # a K a & ' 2 2 3 3 = $ % = S GMSun K s is independent of the mass of the planet, and so is valid for any planet
Kepler s Third Law, final If an object is orbiting another object, the value of K will depend on the object being orbited For example, for the Moon around the Earth, K Sun is replaced with K Earth
Back to inverse square law: Finding the Value of G In 1789 Henry Cavendish measured G The two masses are fixed at the ends of a light horizontal rod Two large masses were placed near the small ones The angle of rotation was measured https://www.youtube.com/watch?v=ym6nlwvqzne
Law of Gravitation, cont This is an example of an inverse square law The magnitude of the force varies as the inverse square of the separation of the particles The law can also be expressed in vector form! F 12 = G m m 1 2 ˆr r 2 12 The negative sign indicates an attractive force
Notation! F is the force exerted by particle 1 on 12 particle 2! F 21 is the force exerted by particle 2 on particle 1 If there is a negative sign in the vector form of the equation, it indicates that particle 2 is attracted toward particle 1
More About Forces! F 12 = F! 21 The forces form a Newton s Third Law action-reaction pair Gravitation is a field force that always exists between two particles, regardless of the medium between them The force decreases rapidly as distance increases A consequence of the inverse square law
Gravitational Force Due to a Distribution of Mass The gravitational force exerted by a finitesize, spherically symmetric mass distribution on a particle outside the distribution is the same as if the entire mass of the distribution were concentrated at the center The force exerted by the Earth on a particle of mass m near the surface of the Earth is F g Mm = G R E 2 E
G vs. g Always distinguish between G and g G is the universal gravitational constant It is the same everywhere g is the acceleration due to gravity g = 9.80 m/s 2 at the surface of the Earth g will vary by location
Finding g from G The magnitude of the force acting on an object of mass m in freefall near the Earth s surface is mg This can be set equal to the force of universal gravitation acting on the object mg M m = G R M g = G R E 2 E E 2 E
g Above the Earth s Surface If an object is some distance h above the Earth s surface, r becomes R E + h g = GM E ( R ) 2 E + h This shows that g decreases with increasing altitude As r, the weight of the object approaches zero
Variation of g with Height
Shell Theorem G Force anywhere inside a spherical shell = 0 Consider any point P with mass m within a circular shell at a distance d from the wall of the shell. P θ θ x d θ θ Force due to the infinitesimally small circular disk could be considered independent of the distance between them x For every small shaded region on one side of the sphere there lies an opposite side that exerts an equal but opposite force, the net force inside any point of the shell = 0
Barycenter a moment of truth Do lighter bodies always revolve around heavier ones? NO. They both revolve around their CoM. In the case of earth and moon, CoM lies inside the earth.
Earth-moon barycenter hkps://www.youtube.com/watch? v=ugbangbrkws Solar System barycenter hkps://www.youtube.com/watch? v=1isr3yw6fxo
Gravitational Potential Energy The gravitational force is conservative The change in gravitational potential energy of a system associated with a given displacement of a member of the system is defined as the negative of the work done by the gravitational force on that member during the displacement Δ U = U = ( ) f Ui F r dr r f r i
Gravitational Potential Energy At infinity we consider the reference U = 0 As a particle moves from A to B, its gravitational potential energy changes by ΔU = -W Going up, up away ΔU is more positive r 2 r 1 Approaching earth ΔU is more negative
Near the earth objects Celestial Objects GMEm Ur ( ) = r This is valid only for r R E and not valid for r < R E
Gravitational Potential Energy for the Earth, cont Graph of the gravitational potential energy U versus r for an object above the Earth s surface The potential energy goes to zero as r approaches infinity
Systems with Three or More Particles The total gravitational potential energy of the system is the sum over all pairs of particles Gravitational potential energy obeys the superposition principle
Systems with Three or More Particles, cont Each pair of particles contributes a term of U Assuming three particles: U = U + U + U total 12 13 23 = mm mm + + mm 1 2 1 3 2 3 G r 12 r 13 r 23 The absolute value of U total represents the work needed to separate the particles by an infinite distance
Energy and Satellite Motion Assume an object of mass m moving with a speed v in the vicinity of a massive object of mass M M >> m Also assume M is at rest in an inertial reference frame The total energy is the sum of the system s kinetic and potential energies
Energy and Satellite Motion, 2 Total energy E = K +U 1 Mm E 2 = 2 mv G r In a bound system, E is necessarily less than 0
A satellite of mass m is in a circular orbit of radius R around Earth ( radius r E, mass M) (a) What is the Total Mechanical Energy ( where U gravity = 0 as R -> infinity) (b) How much work would be required to move the satellite into a new orbit with radius 2R a) mv 2 /R = GMm/R 2 K = ½ mv 2 = > GMm/(2R) TME = U + K = -GMm/R + GMm/(2R) = -GMm/(2R) b) K i + U i +W = K f +U f W = E f - E i = E at 2R E at R = -GMm/(2*2R) (-GMm/(2R)) = GMm/(4R)
Energy in a Circular Orbit An object of mass m is moving in a circular orbit about M The gravitational force supplies a centripetal force GMm E = 2r
Energy in an Elliptical Orbit For an elliptical orbit, the radius is replaced by the semimajor axis E GMm = 2a The total mechanical energy is negative The total energy is constant if the system is isolated
Escape Velocity K i + U i = K f +U f To get far away from earth means U f =0 To find minimum K i, we will assume minimum value of K f = 0 K i + U i = 0 => K i = -U i To escape earth one needs
Escape Speed, General The Earth s result can be extended to any planet v esc = 2GM R The table at right gives some escape speeds from various objects
Escape Speed, Implications Complete escape from an object is not really possible The gravitational field is infinite and so some gravitational force will always be felt no matter how far away you can get This explains why some planets have atmospheres and others do not Lighter molecules have higher average speeds and are more likely to reach escape speeds
What is the gravitational force on a particle of mass m at a distance x from the center of a spherically symmetric planet of uniform density p, total mass M and radius R for x R x < R Ans a) GMm/x 2 for x R b) GMm x/r 3 for x < R
Given that the radius of earth is 6.37 * 10 6, determine the mass of earth. Ans: 6.0 X 10 24 Kg
Communication satellites are often parked in geosynchronous orbits above Earth s surface. Such satellites have orbit periods equal to earth s rotation period., so they remain above the same position on Earth s surface. Determine the altitude and speed that a satellite must have to be in geosynchronous orbit above a fixed point on Earth s equator (M earth = 5.98 X 10 24 Kg)
A uniform, slender bar of mass M has length L. Determine the gravitational force it exerts on the point of mass m shown below: & & && & 8z& x& dm& L& & & dx& Ans:&F&=&GMm/z(z+L)&