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Orbital Motion Gravitation provides the centripetal force for circular orbital motion The behaviour of the solar system is summarised by Kepler´s laws Kepler´s law state 1. Each planet moves in an ellipse which has the sun at one focus 2. The line joining the sun to the moving planet sweeps out equal areas in equal times
Deriving the Third Law Suppose a planet of mass m moves with speed v in a circle of radius r round the sun of mass M The gravitational attraction of the sun for the planet is = G Mm r2 From Newton’s Law of Universal Gravitation
If this is the centripetal force keepingthe planet in orbit thenG Mm = mv2 (from centripetal equation) r2 r∴ GM = v2 r
If T is the time for the planet to make oneorbitv = 2π r v2 = 22π2 r2 T T2∴ GM = 4π 2 r 2 r T2∴ GM = 4π 2 r 3 T2∴ r 3 = GM T2 4π 2∴ r 3 = a constant T2
Kepler´s Third Law The square of the times of revolution of the planets (i.e. Their periodic time T ) about the sun are proportional to the cubes of their mean distance (r) from it. We have considered a circular orbit but more advanced mathematics gives the same result for an elliptical one.
Energy of Orbiting Satellites Potential Energy, Ep A satellite of mass m orbiting the Earth at a distance r from its centre has Gravitational Potential, V = - G Me r Therefore the gravitational potential energy, E p = - G M em r
Energy of Orbiting Satellite Kinetic Energy, Ek By the law of Universal Gravitation and Newton’s Second Law G Mem = mv2 r2 r Therefore the kinetic energy, E k = ½mv 2 = G M e m 2r
Energy of Orbiting Satellite Total Energy, Ep+ Ek Total Energy = - G Mem + G Mem r 2r Total Energy = - G M e m 2r Total energy is constant for a circular orbit.
Graphs of Energy AgainstRadius Potential Energy Ep Ep r 1/r i.e. Ep ∝ -1/r or Ep = -k/r where k is the constant of proportionality = GMem
Graphs of Energy AgainstRadius Kinetic Energy Ek Ek r 1/r i.e. Ek ∝ 1/r or Ek = k/r where k is the constant of proportionality = GMem 2
Graphs of Energy AgainstRadius Total Energy Etotal Etotal r 1/r i.e. Etotal ∝ -1/r or Etotal = -k/r where k is the constant of proportionality = GMem 2
Weightlessness Consider an object of mass m hanging from a spring balance which is itself hanging from the roof of a lift T a mg
The body is subjected to a downward directedforce mg due to the Earth, and an upwardsforce T, due to the tension in the spring.The net down ward force is (mg – T ), and byNewton´s second law mg – T = maWhere a is downward directed acceleration ofthe body
If the lift is stationary, or is moving with aconstant speed, a = 0 and therefore T = mgi.e. The balance registers the weight of thebody as mgHowever, if the lift is falling freely undergravity, both it and the body have adownward directed acceleration of gi.e. g = a
It follows from the equation that T = 0i.e. The balance registers the weight of thebody as zeroIt is usual to refer to a body in this situation asbeing weightlessThe term should be used with care, agravitational pull of magnitude mg acts on thebody whether it is in free fall or not, andtherefore, in the strictest sense it has weighteven when in free fall.
The reason it is said to be weightless is that,whilst falling freely, it exerts no force on itssupport.Similarly, a man standing on the floor of a liftwould exert no force on the floor if the liftwere in free fall. In accordance with Newton´sthird law, the floor of the lift would exert noupwards push on the man and therefore hewould not have the sensation of weight.
Orbital Weightlessness An astronaut in an orbiting spacecraft has a centripetal acceleration equal to g1, where g1 is the acceleration due to gravity at the height of the orbit The spacecraft also has the same centripetal acceleration The astronaut therefore has no acceleration relative to his spacecraft, i.e. he is weightless