Gravitation © David Hoult 2009.

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Presentation transcript:

Gravitation © David Hoult 2009

© David Hoult 2009

© David Hoult 2009

© David Hoult 2009

© David Hoult 2009

© David Hoult 2009

© David Hoult 2009

F a m1m2 © David Hoult 2009

© David Hoult 2009

© David Hoult 2009

© David Hoult 2009

© David Hoult 2009

© David Hoult 2009

© David Hoult 2009

1 F a r2 © David Hoult 2009

m1m2 F = G r2 where G is the universal gravitation constant © David Hoult 2009

m1m2 F = G r2 N m2 kg-2 where G is the universal gravitation constant © David Hoult 2009

Testing the Inverse Square Law of Gravitation © David Hoult 2009

© David Hoult 2009

9.8 = 2.72 × 10-3 ms-2 602 © David Hoult 2009

9.8 = 2.72 × 10-3 ms-2 602 v2 a = r © David Hoult 2009

9.8 = 2.72 × 10-3 ms-2 602 v2 a = r r = 3.84 × 108 m T = 27.3 days © David Hoult 2009

Centripetal acceleration of the moon (caused by the force of gravity) 2.72 × 10-3 ms-2 © David Hoult 2009

The inverse square law is a good theory Conclusion The inverse square law is a good theory © David Hoult 2009

Relation between g and G © David Hoult 2009

Relation between g and G © David Hoult 2009

Relation between g and G © David Hoult 2009

Relation between g and G © David Hoult 2009

we have assumed the equivalence of inertial and gravitational mass © David Hoult 2009

Gravitational Field Strength © David Hoult 2009

The g.f.s. at a point in a gravitational field is the force per unit mass acting on point mass © David Hoult 2009

The g.f.s. at a point in a gravitational field is the force per unit mass acting on point mass Units Nkg-1 © David Hoult 2009

“Force per unit mass” is equivalent to acceleration © David Hoult 2009

G.f.s. is another name for acceleration due to gravity © David Hoult 2009

© David Hoult 2009

© David Hoult 2009

1 g a r2 © David Hoult 2009

1 g a r2 © David Hoult 2009

© David Hoult 2009

1 outside the sphere g a r2

1 outside the sphere g a r2 © David Hoult 2009

1 outside the sphere g a r2 inside the sphere g a r © David Hoult 2009

1 outside the sphere g a r2 inside the sphere g a r © David Hoult 2009

© David Hoult 2009

© David Hoult 2009

World High Jump Record... © David Hoult 2009

World High Jump Record... on Mars ? © David Hoult 2009

© David Hoult 2009

© David Hoult 2009

maximum height, s depends on: © David Hoult 2009

maximum height, s depends on: initial velocity, u © David Hoult 2009

maximum height, s depends on: initial velocity, u acceleration due to gravity, g © David Hoult 2009

so, for a given initial velocity u2 = -2gs so, for a given initial velocity © David Hoult 2009

so, for a given initial velocity u2 = -2gs so, for a given initial velocity gs = a constant © David Hoult 2009

For a given initial velocity, the maximum height reached by the body is inversely proportional to the acceleration due to gravity © David Hoult 2009

1 s a g © David Hoult 2009

1 s a g sg = a constant © David Hoult 2009

1 s a g gs = a constant g1s1 = g2s2 or s1 g2 = s2 g1 © David Hoult 2009

Gravitational Potential © David Hoult 2009

The potential at a point in a gravitational field is the work done per unit mass moving point mass from infinity to that point © David Hoult 2009

units of potential J kg-1 The potential at a point in a gravitational field is the work done per unit mass moving point mass from infinity to that point units of potential J kg-1 © David Hoult 2009

© David Hoult 2009

© David Hoult 2009

© David Hoult 2009

w = Fs but in this situation the force is not of constant magnitude © David Hoult 2009

w = Fs but in this situation the force is not of constant magnitude © David Hoult 2009

It is clear that the work done will depend on: © David Hoult 2009

It is clear that the work done will depend on: the mass of the planet, M © David Hoult 2009

It is clear that the work done will depend on: the mass of the planet, M the distance, r of point p from the planet © David Hoult 2009

It is clear that the work done will depend on: the mass of the planet, M guess: w a M the distance, r of point p from the planet © David Hoult 2009

It is clear that the work done will depend on: the mass of the planet, M guess: w a M the distance, r of point p from the planet guess: w a 1/r © David Hoult 2009

...it can be shown that... © David Hoult 2009

GM w = r © David Hoult 2009

“at infinity” means that the body is out of the gravitational field A body at infinity, has zero gravitational potential “at infinity” means that the body is out of the gravitational field © David Hoult 2009

All bodies fall to their lowest state of potential (energy) A body at infinity, has zero gravitational potential “at infinity” means that the body is out of the gravitational field All bodies fall to their lowest state of potential (energy) © David Hoult 2009

A body at infinity, has zero gravitational potential “at infinity” means that the body is out of the gravitational field All bodies fall to their lowest state of potential (energy) All gravitational potentials are therefore negative quantities © David Hoult 2009

GM V = r © David Hoult 2009

GM V = r Therefore the gravitational potential energy possessed by a body of mass m placed at point p is given by © David Hoult 2009

GM V = r The gravitational potential energy possessed by a body of mass m placed at point p is given by G P E = V m © David Hoult 2009

Escape Velocity © David Hoult 2009

© David Hoult 2009

© David Hoult 2009

© David Hoult 2009

© David Hoult 2009

G P E = zero © David Hoult 2009

G P E = zero To find the minimum velocity, ve which will cause the rocket to escape the Earth’s gravity, assume K E of distant rocket is also equal to zero. © David Hoult 2009

G P E = zero To find the minimum velocity, ve which will cause the rocket to escape the Earth’s gravity, assume K E of distant rocket is also equal to zero. As the body is moving away from the planet it is losing K E and gaining G P E © David Hoult 2009

G P E = zero To find the minimum velocity, ve which will cause the rocket to escape the Earth’s gravity, assume K E of distant rocket is also equal to zero. As the body is moving away from the planet it is losing K E and gaining G P E D K E = D G P E © David Hoult 2009

If the mass of the rocket is m, then the G P E it possesses at the surface of the planet is GMm G P E = R © David Hoult 2009

If the mass of the rocket is m, then the G P E it possesses at the surface of the planet is GMm G P E = R GMm D G P E = r © David Hoult 2009

GMm G P E = R GMm D G P E = R ½mve2 If the mass of the rocket is m, then the G P E it possesses at the surface of the planet is GMm G P E = R GMm D G P E = R D K E = ½mve2 © David Hoult 2009

GMm ½mve2 = R © David Hoult 2009

Also, as g = GM/R2 © David Hoult 2009

Also, as g = GM/R2 © David Hoult 2009