Physics 01, Lecture Today s Topics n Universal Gravitation (Chapter 1 n Review: Newton s Law of Universal Gravitation n Properties of Gravitational Field (1.4 n Gravitational Potential Energy (1.5 n Escape Speed (1.6 n Hope you ve previewed chapter 1. by 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 F 1 = G m m 1 ˆr r 1 on F 1 = G m 1m ˆr r 1 = F 1 Universal Gravitational Constant: G = 6.67x10-11 Nm /kg m F 1 =-F 1 F 1 F 1 r 1 r 1 = r 1 ˆr 1 m 1 Review: G and g q Gravity on earth (free fall gravity is just one example of the Universal Gravity g q Above earth surface: g As Function of Height m F g = GMm/r = m( GM/r On Earth: g= GM E / g= GM E /( +h M Also: At Equator: =6.78x10 6 m, g=9.780 At the Poles: =6.57x10 6 m, g=9.8 1
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 Phy0, conceptual only for now The gravitational field exerts a gravitational force on another mass (any mass in the field. M m g = GM r ˆ r F g = g m = GMm r ˆ r Properties of the Gravitational Force q It is very weak! G = 6.67x10-11 Nm /kg 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 : 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 r f Gm 1 m r i = ( U f ( U i Gm1m = r U Gravitational Potential Energy Gm1m = r U g 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 /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. we take r= as reference point for U g =0 q see the math: At r =, U g ( = GM E m at r = + h, U g ( + h = GM E m + h Take r = as reference for U = 0 U g = U g ( + h U g ( = GM Em + h ( GM h<< Em = m GM E h g = GM E
Gravitational Potential Energy: System with 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 1 1! m1m m1m mm " = G $ + + % & r 1 r 1 r ' Earth as Inertia Reference Frame q Inertial Frame: Where Newton s (1 st and 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 = 0.0059 m/s (effects from other stellar objects smaller ü In reality, the Earth does spin a c = ω = 6.7x10 6 x (π/86400 = 0.005 m/s (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. 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. CM Orbiting Escaping Even masses M>>m
Dynamics of Orbiting q For a circular orbit: Force on m: F G = GMm/r Orbiting speed: v = GM/r (independent of m Kinetic energy: KE= ½ mv = ½ GMm/r Potential energy U G = = - GMm/r (note: U G = - KE! Total Energy: E = KE + U G = - ½ GMm/r (=-KE Satellite: Orbiting Period q Newton s nd Law with gravitational force: mv /r = GM E m/r à v = (GM E /r ½ = (GM E /( +h ½ à period T = πr/v = π r / / (GM E ½ Note: r = + h è for low orbit h <<, r ~ T ~ π (RE+h / / (GM E ½ ~ 85-10 minutes q For a generic elliptical orbit : It can be derived (not required: E = - ½ GMm/a (a: semi-major axis Exercise: Geosynchronous Satellite q What is the height of a geostationary/geosynchronous satellite? Ø Newton s nd Law with gravitational force: mv /r = GM E m/r à v = (GM E /r ½ = (GM E /( +h ½ à period T = πr/v = π r / / (GM E ½ For geostationary satellite: T = 1day = 86400 s Answer: r = 4164 km, h=r- = 5786 km Exercise: Energy Required to Change Orbit of a Satellite q A 470 kg communication satellite, initially at orbit of h i =80 km, is to be boosted to the geosynchronous orbit of h f = 5786 km, Ø What is the energy required to do so? Ø Solution: At 80km, E i = - ½ GM E m /r i = -½ GM E m/( +80km = - 1.41x10 10 J At 5786km, E f = - ½ GM E m /r i = -½ GM E m/( +5786km = -0.x10 10 J Geostationary orbit for communication satellite Energy required = E f E i = 1.19x10 10 J 4
Escape Speed q Escape: h = E = KE + PE =0 q At the Earth s surface (r i = KE = ½ mv esc PE = (-GM E m/ E = ½ mv esc + (-GM E m/ q Energy conservation: E= E = 0 ½ mv esc + (-GM E m/ =0 à work out algebra after class. 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 = 11. km/s q Third Astronautical Speed: (Escaping Solar system v = 16.7 km/s v esc = GM E =11. km/s (Escape Speed 5