# Handout III : Gravitation and Circular Motion

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Handout III : Gravitation and Circular Motion
EE1 Particle Kinematics : Newton’s Legacy “I frame no hypotheses; for whatever is not deduced from the phenomena is to be called a hypothesis; and hypotheses, whether metaphysical or physical, whether of occult qualities or mechanical, have no place in experimental philosophy.” Chris Parkes October

Gravitational Force Myth of Newton & apple. He realised gravity is universal same for planets and apples Any two masses m1,m2 attract each other with a gravitational force: F F m1 m2 r Newton’s law of Gravity Inverse square law 1/r2, r distance between masses The gravitational constant G = 6.67 x Nm2/kg2 Explains motion of planets, moons and tides mE=5.97x1024kg, RE=6378km Mass, radius of earth Gravity on earth’s surface Or Hence,

Circular Motion Rotate in circle with constant angular speed 
360o = 2 radians 180o =  radians 90o = /2 radians Circular Motion Rotate in circle with constant angular speed  R – radius of circle s – distance moved along circumference =t, angle  (radians) = s/R Co-ordinates x= R cos  = R cos t y= R sin  = R sin t Velocity =t R s y t=0 x Acceleration

Magnitude and direction of motion
Velocity v=R And direction of velocity vector v Is tangential to the circle v Acceleration a a= 2R=(R)2/R=v2/R And direction of acceleration vector a a= -2r Acceleration is towards centre of circle

I is called moment of inertia
Angular Momentum For a body moving in a circle of radius r at speed v, the angular momentum is L=(mv)r = mr2= I  The rate of change of angular momentum is The product rF is called the torque of the Force Work done by force is Fs =(Fr)(s/r) = Torque  angle in radians Power = rate of doing work = Torque  Angular velocity (using v=R) I is called moment of inertia s r

Force towards centre of circle
Particle is accelerating So must be a Force Accelerating towards centre of circle So force is towards centre of circle F=ma= mv2/R in direction –r or using unit vector Examples of central Force Tension in a rope Banked Corner Gravity acting on a satellite

Satellites Centripetal Force provided by Gravity m R M
N.B. general solution is an ellipse not a circle - planets travel in ellipses around sun Satellites Centripetal Force provided by Gravity m R M Distance in one revolution s = 2R, in time period T, v=s/T T2R3 , Kepler’s 3rd Law Special case of satellites – Geostationary orbit Stay above same point on earth T=24 hours

Gravitational Potential Energy
How much work must we do to move m1 from r to infinity ? When distance R Work done in moving dR dW=FdR Total work done r m1 m2 Choose Potential energy (PE) to be zero when at infinity Then stored energy when at r is –W -ve as attractive force, so PE must be maximal at 

Compare Gravitational P.E.
Relate to other expression that you know Potential Energy falling distance h to earth’s surface = mgh Uses: Expression for g from earlier g=GME/RE2 Binomial expansion given h<<RE (1+)-1 = 1- +…..smaller terms… Compare with Electrostatics: Same form, but watch signs: attractive or repulsive force attract repel Maximal at  Minimal at

A final complication: what do we mean by mass ?
Newton’s 2nd law F = mI a Law of Gravity mI is inertial mass mG, MG is gravitational mass - like electric charge for gravity Are these the same ? Yes, but that took another 250 years till Einstein’s theory of relativity to explain!