9: Motion in Fields

9: Motion in Fields 9.2 Gravitational field, potential and energy

Gravity Recap: Newton’s universal law of gravitation: Gravitational field strength: F = GMm r2 …the force per unit mass experienced by a small test mass (m) placed in the field. g = GM r2

GPE in a uniform field When we do vertical work on a book, lifting it onto a shelf, we increase its gravitational potential energy (Ep). If the field is uniform (e.g. Only for very short distances above the surface of the Earth) we can say... GPE gained (Ep) = Work done = F x d = Weight x Change in height so...ΔEp = mg∆h E.g. In many projectile motion questions we assume the gravitational field strength (g) is constant.

GPE in non-uniform fields However, as Newton’s universal theory of gravity says, the force between two masses is not constant if their separation changes significantly. Also, the true zero of GPE is arbitrarily taken not as Earth’s surface but at ‘infinity’. Ep = 0 ‘Infinity’ If work must be done to “lift” a small mass from near Earth to zero at infinity then at all points GPE must be negative. (This is not the same as change in GPE which can be + or -) Lots of positive work must be done on the small mass! Ep = negative

The gravitational potential energy of a mass at any point is defined as the work done in moving the mass from infinity to that point. The GPE of any mass will always be due to another mass (after all, what is attracting it from infinity?) Strictly speaking, the GPE is thus a property of the two masses. E.g. Calculate the potential energy of a 5kg mass at a point 200km above the surface of Earth. ( G = 6.67  10-11 N m2 kg-2 , mE= 6.0  1024 kg, rE= 6.4  106 m ) Ep = - GMm r

The gravitational potential energy of a mass at any point is defined as the work done in moving the mass from infinity to that point.

a b Q. What do the indicated properties of these two graphs represent?

Gravitational Potential Whereas gravitational force on an object on Earth depends upon the mass of the object itself, gravitational field strength is a measure of the force per unit mass of an object at a point in Earth’s field. Similarly, whereas the GPE of say a satellite, depends upon both the mass of Earth and the satellite itself, gravitational potential is a measure of the energy per unit mass at a point in Earth’s field.

Thus for a field due to a (point or spherical) mass M: So ... E.g. Calculate the potential of a 5kg mass at a point 200km above the surface of Earth. What would be the potential of a 10kg mass at the same point? ( G = 6.67  10-11 N m2 kg-2 , mE= 6.0  1024 kg, rE= 6.4  106 m ) The gravitational potential at a point in a field is defined as the work done per unit mass in bringing a point mass from infinity to the point in the field. V = Ep = - GMm m r m V = Gravitational potential (Jkg-1) V = - GM r

Gravitational Potential in a uniform field. For a uniform field… ∆Ep = mg∆h So… ∆V = ∆Ep = mg∆h m m ∆V = g∆h

How far apart are the equipotentials in this diagram?

Equipotential Surfaces Equipotential surfaces or lines join points of equal potential together. Thus if a mass is moved around on an equipotential surface no work is done. Thus the force due to the field, and therefore the direction of the field lines, must be perpendicular to the equipotential surfaces at all times.

Potential Gradient • The separation of the equipotential surfaces tells you about the field: • Uniform fields have equal separation • Fields with decreasing field strength have increasing separation.

If the equipotentials are close together, a lot of work must be done over a relatively short distance to move a mass from one point to another against the field – i.e. the field is very strong. This gives rise to the concept of ‘potential gradient’. The ‘potential gradient’ is given by the formula... Potential gradient = ΔV Δr It is related to gravitational field strength... g = - ΔV Δr

Escape speed If a ball is thrown upwards, Earth’s gravitational field does work against it, slowing it down. To fully escape from Earth’s field, the ball must be given enough kinetic energy to enable it to reach infinity. Loss of KE = Gain in GPE ½ mv2 = GMm (Note this also = Vm) r So... but… so… The escape speed is the minimum launch speed needed for a body to escape from the gravitational field of a larger body (i.e. to move to infinity).

Note we could also say... ½ mv2 = GMm = Vm r So... v = √(2V)

Note we could also say... • ½ mv2 = GMm = Vm • r • so... v = √(2V) • Assumptions… • Planet is a perfect sphere • No other forces other than gravitational attraction of the planet. • Note: • Applies only to projectiles • - Direction of projection is not important if we assume that the planet is not rotating

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9: Motion in Fields