Physics 201, Lecture 23 Today s Topics n Universal Gravitation (Chapter 13) n Review: Newton s Law of Universal Gravitation n Properties of Gravitational Field (13.4) n Gravitational Potential Energy (13.5) n Escape Speed (13.6) n Hope you ve previewed chapter 13.
Review: Newton s Law Of Universal Gravitation q The Law: Any pair of objects in the Universe attract each other with a force that is proportional to the products of their masses and inversely proportional to the square of their distance by F = G m 1 m 2 ˆr 12 r 2 12 on F = G m 1m 2 ˆr 21 r 2 = F 21 12 m 2 F 12 =-F 21 F 12 F 21 r 12 m 1 Universal Gravitational Constant: G = 6.673x10-11 Nm 2 /kg 2 r 12 = r 12 ˆr 12
Review: G and g q Gravity on earth (free fall gravity) is just one example of the Universal Gravity g F g = GMm/r 2 = m( GM/r 2 ) m On Earth: g= GM E /R E 2 M
q Above earth surface: g As Function of Height g= GM E /(R E +h) 2 Also: At Equator: R E =6.378x10 6 m, g=9.780 At the Poles: R E =6.357x10 6 m, g=9.832
The Gravitational Field q In modern physics, the gravitational force can be viewed as: A mass (any mass) creates a gravitational field around it. Field: A physical quantity filled up in space (More on field in Phy202, conceptual only for now) The gravitational field exerts a gravitational force on another mass (any mass) in the field. m g = GM r 2 ˆ r M F g = g m = GMm r 2 ˆ r
Properties of the Gravitational Force q It is very weak! G = 6.673x10-11 Nm 2 /kg 2 e.g Two 1 Kg particles, 1 meter away F g = 6.67x10-11 N q It is one of four fundamental forces: (Strong, Electromagnetic, weak, gravity) q It is proportional to 1/r 2 : double r ¼ F G q It is a conservative force. (Work independent of path) W = r i r f F G d r = A potential energy can be defined. Gm 1 m 2 r f Gm 1m 2 r i = ( U f ) ( U i ) Gm1m r 2 U =
Gravitational Potential Energy Gm m U = 1 2 g r Ø It is negative. (but that s no problem) Ø U g =0 if r = (i.e. We use infinity as reference point for U g =0)
What is the mgh We have Being Talking About q Now we have encountered two forms of gravitational potential energy Earlier in the semester: U g =mgh Today : U g = - Gm 1 m 2 /r q How do they compare? Ø the mgh form is a special case under conditions: 1. the object m is on or near earth s surface 2. we take r=r E as reference point for U g =0 q see the math: At r = R E, U g (R E ) = GM E m R E at r = R E + h, U g (R E + h) = GM E m R E + h Take r = R E as reference for U = 0 U g = U g (R E + h) U g (R E ) = GM m E R E + h ( GM m h<<r E E ) R E = m GM E 2 R E h g = GM E R E 2
Gravitational Potential Energy: System with 3 Or More Masses q For a system with three or more masses, the total gravitational potential energy is the sum over all pair of masses U = U + U + U total 12 13 23! m1m 2 m1m 3 m2m " 3 = G $ + + % & r 12 r 13 r 23 '
Earth as Inertia Reference Frame q Inertial Frame: Where Newton s (1 st and 2 nd ) laws are valid Ø No force, no acceleration. Ø To be an inertia reference frame, the Earth shall be subject to no force have no spin. ü In reality, the Earth does subject to gravitational force from the Sun (and other stars/planets/satellites) g at_earth_by Sun = G M Sun /R Sun_Earth 2 = 0.0059 m/s 2 (effects from other stellar objects smaller) ü In reality, the Earth does spin a c = R E ω 2 = 6.37x10 6 x (2π/86400) 2 = 0.0053 m/s 2 (at Equator) Quick quiz: What about directions for the above accelerations?
Motion of Two Star System q For a two-star system, the path of each star is an elliptical orbit around the CM of the system. If one star is much heavier than the other, then it seems that the lighter star is orbiting around the heavier one. CM Even masses M>>m
Orbiting and Escaping q For a small object m in in motion near a massive object M, under sole influence of gravity, one can take the approximation: the big object M is considered stationary the smaller object m is either in elliptical (orbiting) or parabolic (escaping/passing-by) path around M. Orbiting Escaping
q For a circular orbit: Dynamics of Orbiting Force on m: F G = GMm/r 2 Orbiting speed: v 2 = GM/r (independent of m) Kinetic energy: KE= ½ mv 2 = ½ GMm/r Potential energy U G = = - GMm/r (note: U G = -2 KE!) Total Energy: E = KE + U G = - ½ GMm/r (=-KE) q For a generic elliptical orbit : It can be derived (not required): E = - ½ GMm/a (a: semi-major axis)
Satellite: Orbiting Period q Newton s 2 nd Law with gravitational force: mv 2 /r = GM E m/r 2 à v = (GM E /r) ½ = (GM E /(R E +h)) ½ à period T = 2πr/v = 2π r 3/2 / (GM E ) ½ Note: r = R E + h è for low orbit h << R E, r ~ R E T ~ 2π (RE+h) 3/2 / (GM E ) ½ ~ 85-120 minutes
Exercise: Geosynchronous Satellite q What is the height of a geostationary/geosynchronous satellite? Ø Newton s 2 nd Law with gravitational force: mv 2 /r = GM E m/r 2 à v = (GM E /r) ½ = (GM E /(R E +h)) ½ à period T = 2πr/v = 2π r 3/2 / (GM E ) ½ For geostationary satellite: T = 1day = 86400 s Answer: r = 42164 km, h=r-r E = 35786 km Geostationary orbit for communication satellite
Exercise: Energy Required to Change Orbit of a Satellite q A 470 kg communication satellite, initially at orbit of h i =280 km, is to be boosted to the geosynchronous orbit of h f = 35786 km, Ø What is the energy required to do so? Ø Solution: At 280km, E i = - ½ GM E m /r i = -½ GM E m/(r E +280km) = - 1.41x10 10 J At 35786km, E f = - ½ GM E m /r i = -½ GM E m/(r E +35786km) = -0.22x10 10 J Energy required = E f E i = 1.19x10 10 J
Escape Speed q Escape: h = E = KE + PE =0 q At the Earth s surface (r i =R E ) KE = ½ mv esc 2 PE = (-GM E m/r E ) E = ½ mv esc 2 + (-GM E m/r E ) q Energy conservation: E= E = 0 ½ mv esc 2 + (-GM E m/r E ) =0 à work out algebra after class. v esc = 2GM E R E =11.2 km/s (Escape Speed)
Three Astronautical Speeds q First Astronautical Speed: (Orbiting near earth surface) v 1 = 7.9 km/s q Second Astronautical Speed: (Escaping the earth) v 2 = 11.2 km/s q Third Astronautical Speed: (Escaping Solar system) v 3 = 16.7 km/s