∆G.P.E = mg∆h When you have previously considered gravitational potential energy, you would have used the equation: Consider a monkey climbing 2 metres up a tree If we were to make the monkey go 50km up towards the edge of space, would this equation still be valid? If that same monkey were to climb to 4 metres high in the next tree its change in potential energy would be double.

We use the idea of field lines to represent both the magnitude and direction of any gravitational field. The field lines show the direction in which a small test mass would move if placed into the field. The higher the density of field lines, the stronger the field. Over small distances (?) the Earths gravitational field can considered to be uniform, that is the field lines are parallel and equal distance apart

The strength of the Earth’s gravitational field decreases as you get further from the centre. This is shown by the density of field lines reducing (i.e. they become further apart). At large distances the equation ∆G.P.E = mg∆h no lon ger holds. This is because the gravitational field strength changes (g < 9.8Nkg -1 ) Radial field

What factors affect the size of the gravitational force between any two masses? Gravity is a force that acts between any two objects that have mass. The mass of the 1 st object (m 1 ) The mass of the 2 nd object (m 2 ) The distance between the two objects (r) G is the constant of proportionality (6.67 × m 3 kg -1 s -2 ) The minus sign indicates that the force is attractive.

Calculate the size of the attractive force between the Earth (5.97×10 24 kg) and the moon (7.35× kg). The moon is on average 384,000km away from the Earth. Estimate the size of the attractive force between these two women sitting on a park bench. G =6.67 × m 3 kg -1 s -2 We can consider that the entire mass of an object is found at its centre of gravity.

The Gravitational field strength at any point is the force per unit mass exerted on a mass placed at that point [g= F/m] (1) Calculate the gravitational field strength on the surface of the Earth given that it has a radius of 6378km and a mass of 5.97×10 24 kg. G =6.67 × m 3 kg -1 s -2 (2) What will be the gravitational field strength due to the Earth (a) on top of mount Everest (8848m)? (b) at the height of the international space station (354km)? 9.79Nkg Nkg Nkg -1

Potential energy of spacecraft 0 Earth Outer Space Mars JupiterSaturn What will happen to the G.P.E of the spacecraft as it journeys through the solar system? For now we will consider it to have 0J of G.P.E on the surface of the Earth.

Potential energy of spacecraft 0 Earth Outer Space MarsJupiterSaturn By convention we take the potential energy of an object to be 0J when it is an infinite distance from all other masses, this also avoids the problems of positive and negative potential energies.

Calculate the change in potential energy when the Voyager probe (720kg) travelled from the Earth’s surface to beyond the edge of the solar system, 16 billion km away (1.6 x10 13 m!) G =6.67 × m 3 kg -1 s -2 m E = 5.97 × kg

We can draw lines that are perpendicular to the field lines on our diagram. These are called equipotential lines. Along any equipotential line an object will have the same gravitational potential energy per unit mass. This quantity E p /m is called the gravitational potential.

The Gravitational potential at any point is equal to the work done in bringing a unit mass from infinity to that point. This is a useful quantity as it is independent of the mass of object in the field, but can be used to calculate the potential energy for any mass at any point in a field. 2x10 5 m 5x10 20 kg (1)Calculate the gravitational potential at a distance 2x10 5 m from a 5x10 20 kg mass? (2) Calculate the potential at a distance of 1x10 20 m? (3) Calculate the change in gravitational potential energy if a 1kg object moved between the points: (4) Calculate ∆GPE if the object has a mass of 49kg

Earlier we defined gravitational potential as the gravitational potential energy per unit mass, therefore V=E p /m or E p =Vm This equation gives the gravitational potential energy of test mass m at a point that is distance r from mass M. When we used the equation G.P.E = mgh we were calculating the change in potential energy, but we get around this by creating a reference point with which all other G.P.E’s are compared.

Force per unit mass Potential energy per unit mass