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Gravitation Beyond the Earth’s Surface

Gravitation Beyond the Earth’s Surface. Engineering 1h WJ Easson http://www.see.ed.ac.uk/~bille. Force of Attraction. r. m 1. m 2. Force is proportional to the mass of the bodies Force is inversely proportional to the distance between them

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Gravitation Beyond the Earth’s Surface

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  1. Gravitation Beyond the Earth’s Surface Engineering 1h WJ Easson http://www.see.ed.ac.uk/~bille

  2. Force of Attraction r m1 m2 • Force is proportional to the mass of the bodies • Force is inversely proportional to the distance between them • G=6.673*10-11 Nm2/kg2 is constant of universal gravitation

  3. FBD m M R v Fg r Earth and Satellite and so For the earth R=6370km =6.37*106m M=5.97* 1024 kg so, at the surface g=9.82ms-2

  4. Typical values of g Note: Moon’s mass not quite negligible in this context

  5. Earth not round! • The Earth is not round and, more than this, its density is uneven. This causes the gravitational pull of the Earth to vary between places depending on various factors, such as mountain masses, underground high-density rock formations, oceans, tides, etc. • The variations in g are small and the units used are therefore changed to mGal (1mGal ~ 10-5ms-2 ~ 10-6g). • How to measure? • Earth-bound methods can be used to determine local variations: spring balance, pendulum. • Satellite orbits are affected.

  6. GRACEGravity Recovery and Climate Experiment

  7. GRACE

  8. GRACEhttp://www.csr.utexas.edu/grace/

  9. Levitation ? mg A balancing force must be applied Go to http://www.ed.ac.uk/~bille to see how it’s done

  10. Circular Motion: Example On a flat road, what is the smallest bend that can be negotiated ‘safely’ by a car of 0.8 tonnes travelling at 60mph? Take s = 0.6. Sketch: FBD (from behind): v r Ff Ff R mg Ff is the centripetal force, due to the friction between the tyres and the road

  11. Circular Motion: Example Without the friction between the tyres and the road, the car would continue in a straight line (literally: go off at a tangent) Vertical forces: R - mg = 0 There is no balancing force, so the car accelerates towards the centre Horizontal forces: Ff = ma At the critical speed: Ff = Fmax = sR = smg This must equal the centripetal acceleration: smg = mv2/r  r = v2/sg r = 120m

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