LECTURE 21 Gravity Introduction: Two events were considered very important during early times: 1. Why everything fll towards the earth 2. Why stars revolve. twas assumed that these two oceurrenees had no connection. Aristotle proposed that because Earth is the mother of everything, it attracts all, other objects to itself. Approximately 2000 years ago, Ptokemy coined the ica of Earth being the centre of the universe and that all celestial objects rotate ‘round it In contrast, in the 1éth century, Nicholas Copernicus proposed that the Sun is the universe's centre and thatthe Earth and other celestial bodies, orbit around it In 17 century, Galileo Galilei took his observations and verified the theory of Nicholas Copernicus using his own telescope. In this lectare we will study the concept of gravity and it’s effect on the objects. ‘Newton's Law of Universal Gravitatio (ua and is directed along the line joining the two bodes. Psting in a constant of proportionality, smn PG a2) [Now lesbo abit careful ofthe direetion of the force. Looking at he fig. 21.1) below, a) Fig. (21.1): Two masses my and ma Exeting force on each oer Fan =Force on m; by mo, Fis =Foree on my by me, Both forees are equal in rmagnitude [Fs|~[Fa|=F , but opposite in direction which can be represented by Newton's Third Law, Fis = -Fi, ‘The Gravitational Constani ‘The gravitational constant G is avery small quantity and needs a very sensitive experiment. ‘An early experiment to find G_ involved suspending two masses and measuring the attractive force. Suppose the distanes between these two masses is) L. From the fig. (212) we can work out an expression forthe gravitational torque: Fig. (212) Finding gravitational constantxF = rF sind the gravitational force between two masses,” will multiply this expression by 2 because there are two masses mi and mp va2( Gully 14) the torsion in a thread. The deflection @ can be measured by observing the beam of the light reflected from the small mirror. In cquilibrium the 1cs balance, sous oe ° Or? mM How to find i? aus) 216 In Lecture No. 16, we leamed that a mass suspended on a string will oscillate ‘and it’s oscillating frequency depends on torque. Suppose the mass is suspended, it will oscillate and x: can be found from observing the period of free oscillations, iF at T=2 i a eel yom @17) with’ ~ “2 . The modern value is @ = 6.67259 x 10° '!Nm?/kg? Mass, Volume and Density of the ‘The magnitude of towards its centre is, depends in the quality of matter, ‘which the Earth attracts a body of mass m 21.8) iron, wood, leather, etc. Objects fecl the force in proportion to their masses. According to Newion’s 2°! law of motion, if a body fall freely, then it will accelerate with gravitational acceleration. So, GmMe Re 19) We measure g, the acceleration due to gravity, as 9.8m/s*. From this we can immediately deduce the Earth's mass by subsiuing Re= 6400 km and G = 6.87259 x 10°" Nn’ /kg*, wf Mee 21.10) — [5.97 x 107% ‘What a remarkable achievement! We can still do more. We can find the volume ofthe Earth as: 4 et Vs = G7RE F=mg = [Los 10m? Hence the density of the Earth is: m Ry = 6400 Am aiiVe pra ~ Me = [5462 kem*) ‘So this is 5.462 times greater than the density of water and tells us that the ‘earth must be quite dense inside. The Gi val Potes ‘The gravitational potential is an important quantity. It is the work done in ‘moving a unit mass from infinity to a given point R, and equals: Vir)= oo (21.12) __oM tr Proof: We need to find the work done in bringing a unit mass from the surface of the Earth to a point at infinity (A point where net force is zero). According to conservation of energy, pe V=—Fdr (21.13) Integrate both sides: . [a =—f, Pom airy 0 v(R) = ou [~ £ (11s) cu|+]” 21.16) @ V(R) = vir) = 2 (21.17) R [Negative indicates that the work is done on the object. This work is now stored in the object in the form of gravitational potential energy. Change in Potential Energy: Using Eq, (21.17), let us calculate the change in potential energy AU when we mise a body of mass to a height above the Earth’s surface. Let Re be the radius of the earth and h+Re is the distance from ” the center of the Earth as shown in the figure. : Hence potential energy AU at height h from the h surface of the earth will be: h+Ry AU =U(Rs +h) U(Re) (2.18) AU=GMm (ee - 1.19) | mm (1 GE 2M (: - (: +t) ) (21.20) ‘Now suppose that the distance is much smaller than the Earth's radius. So, for h<< Re, we can.use binomial approximation’. "ie prea arse ee ‘hen, and ane rel numbers anda} <1*) (21.21) ‘Sowe find, (21.22) (21.23) (2124 on for potential energy and the law of conservation of | ‘energy to find the minimum velocity needed for a body to escape the Earth's ‘gravity. Far away from the Earth, the potential energy is zero, and the smallest ‘value forthe kinetic energy is zero. Requiring that: pace? (KE+PB) n= (KBYP.B)g 2125) Gives, Lig - GMm (21.26) gm = 040 21.26) From this, 2 | Pear 21.27) Re = V0Re 21.28) Putting in some numbers we find that for the Earth, ‘YyeI1.2km/s and for the Sun, ve= 618kmis. For a Black Hole, the escape velocity is so high that nothing can escape, even if it could move with the speed of light! (Nevertheless, Black Holes can be observed because when matter falls into them, a certain kind of radiation is emitted) Put vire This value of R is known es Schwarzchild radius of a black hole. Ifa photon (particle of light) is put into the Schwarzchild radius. it will not be able to ‘escape and ultimately it will fal into the black hole. It is interesting to note that the Schwarzchild radius is outside the black hole. img _ GM 129 (21.30)IF Ris the Earth's radius, ond his the height of the satellite above the ground, then r= Ry + h. Hence, 13) For << Rp, we can approxi cM =f St 132) “ \ Re — om 2133) = Vales g= Re ‘We can easily calculate the time for one complete revolution, pat (2134) = 2135) 2136) 2137) voir ‘This gives the important result, cbserved by Kepler nearly 3 conturis ag that, ras 2138) ‘Total ‘What is the total energy of a satellite moving in a circular orbit around the earth? Clearly, it has two parts, kinetic and potential. Remember that the potential energy is negative. So, B=K.E+P.B (21.39) 1 GMgm = omg - (21.40) aM. But,” ras we saw earlier and therefore, ip LeMm. GMm ean 1GMm 1.42) ‘Note that the magnitude of the potential energy is larger than the kinetic ‘energy. [fit wasn, the satelite would not be bound to the Earth! Te negative sign indicates that the work has to be done on the satellite when taken from a point on the Earth toa point at infinity. Geostationary Satell ‘Goo means “Earth” and stationary means “at rest”. A satellite which revolve around the earth with rotation frequency equal to that of the Earth, (1 rotation =24 hn). It appears to be stationary with respect to the Earth. [Radius (the distance from center ofthe Earth) of this satellite can be found:pts 21.43) GM [Rearranging to find. Multiply and divide by R fe (%) RT? (21.44) R) a2 =f (21.45) ae Taking eube root on bath sides: of RT? Va i [42.33 x 10°m| Fig, (21.3): Satellite at Height of the setllite from the surface of earth: fa distance r from the 7 R= 35.93 x 10'm centre of the Earth 6) r Kepler's Laws: German astronomer Johannes Kepler deduce a mathematical model for the motion of the planets. Kepler’s complete analysis of planetary motion is summarized in three laws known as Kepler's laws. + Kepler’s 1 Law: Itstates that: All planets move in elliptical orbits with the Sun at one focus. Fig, (21.4): Motion of planets in elliptical orbit Itisalso know as Law of Orbits. © Kepler’s 2 Law: ‘A famous discovery of the astronomer Johann Kepler some 300 years ago says that, the line joining a planet to the Sun sweeps out equal areas in equal intervals of time. We caneasily see this fom the conservation of angular momentum, Call AA the area swept out in time . Then from the figure we can see that, a= r(r20) (2147 Divide this by At and then take the limit where it becomes very small. dA AA a ar e148) 46 ho a re ( im 22) 14s) atte (21.50) Recall from Lecture No. 13, Angularist) momentum? is the momentum multiplied by the perpendicular distance. aa. a aire =3r) (v=rw) = Fame (Multiply and Divide by m) “Ebr om -- (Lar) Since L is a constant, let's see why. Angular momentum changes only when torque acts on it, Remember from previous lectures that ser a, de Ifr=O then dé and L is a constant. 7 = 0 because the force acting on the Sun duc to the planet, is central means it is acting along the radial direction towards the center of the Sun, hence: og (oe )e “ rx P= x =0 We have proved one of Kepler's laws (with so litle effort)! + Kepler’s 3" Law: ‘Also known as Law of Periods. It states that: The square of the period of revolution of any planet is proportional to the cube of the planet (21.52) distance from the Sun. Lets prove this result for circular orbits. Proof: ‘Suppose a planet orbits around the sun. The gravitational force acting towards the center of the sun will be equal to the centripetal force? acting on the planet. Hence: Rak Mm _ a Qn r Where? TF om _ (20)? 2153) Sh) noe 2 (4?) r (Sir)" (21.54) Example: ‘Suppose a planet revolves around a sun in elliptical orbit. If it has a speed vi at 2 distance di from the Sun. What will be its speed at distance ds? Solution: Using the law of conservation of angular momentum: * Angular momentum: > Centrpetl force:moydy = mvs (21.55) vrdy u= 1.56) ‘t 1 hh (21.56) ectgeonne RO ED energy: GMm | Lig 1s7) ‘Substituting v; in above equation to find v:, _GMm 1, (dh)? __GMm | 1g 1 vd ot Mm GMm ae a & Using a* — * = (a + 6)(a~ 6) in denominator: 2 did: — ds) 20M (Taig may) 2 4 ) verom (4 ° (aa ay) ‘Substituting value in Eq. (21.95) mods = murd, 2GMad "Vd dy) =n, [2OMisds Vianay is a non-impact collision used to give an extra boost to spacecraft. This collision is one dimensional clastic collision.Figure shows a planet with mass M and velocity V. A stallite of mass m moving with velocity vi is direeted to pass near the Planet to increase it’s velocity to ve. Such that M>>m and V- remains constant. Energy and momentum remain conserved during clastic collision in one dimension. In Lecture No. 10, we already proved that the relative velocity before collision is equal to the relative velocity afte collision 4+ V=—(y + ¥) 1.59) vp = (vt WV) (21.60) I is important to note that “A uniformly dense spherical shell attracts an extemal point mass as if the mass of the shell were concentrated at its center”, as show in the figure below. M + Gravitational effect of spherical distribution of matter: “ ‘Another important point to note is, “A uniform spherical shell of matter exerts no gravitational force on a particle located inside it” In other ‘words, the sum of all the forces acting on a mass inside a uniform spherical shel + Atthe Surface of the Earth: rhe value of g at the surface of the earth can be found using Newton's law of universal gravitation. The magnitude of force with which Earth attracts the body towards its centre 161 — (21.62) the surface for a perfectly spherical Earth because it is inversely proportional to Rr?. + Atsome height h above the surf ‘The magnitude of force with which Earth attracts the body towards its centre is:GMam —_ (21.63) The Ra? ere According to Newton’s second law: ne Fomm . Mem (21.64) OOS OE Ra = SMe 21.65) Ge ReF From here, we can find the effective valuc of g: GM a= ee (21.66) -— 167 ECL + ae) (at) =e (21.68) (ra) oe 1 (21.68) (er por BCR ore cet | mos(t + $) (21.70) =a(0 EB) amy Forh=0,g,=9 ‘As discuused in the section of the sling shot effect, when a body of mass m is ‘on the surface or above the surface of the Earth, the Earth works like a point particle, as iPits total mass is concentrated at its centre + Below the surface of the Earth: ‘As the particle lies inside the shell of radius d, there is no force on the particle ‘due to the shell. The only force exerted comes from the sphere of radius Re — . The mass of this smaller sphere is: Mb= pe (Gace - a) 17), 5 Sur the density ofthis phere is: Mp ee e173) OR) Now we can find out the effective ‘acceleration due to gravity. The force ‘acting on the mass m will be:GMpn Pome Rea? (21.74) G loz (57(Rs —@*)] 4 1.75) a Sr (Substitute Eq. (21.72 (21-75) = Gor (Fite ») (21.76) “(Take Re common) (21.76) -; (Multiply and Divide by Ri) +) (Substitute Mp from Eq, (21.73)) From the above equation, we realize that $4=9 when d=0. If we increase d then gq decreases which implies that effective gravitational acceleration decreases as we go below the surface of the Earth, At the center of the | Dia he net ath Earth where d = Re the value ‘Varatn ofa proibtionl coteton of gq = 0. Hence there is 0 que tothe Barth with distance trom the centre ofthe Barth gravitational force at the center of the Earth as shown in the figure. ROR aR