Gravitational potential Is this the same as gravitational potential energy?
F = g = F m g = (-) GM R 2 g x = 4 3 πgρx Review GMm R2 force b/w two point masses gravitational field strength on Earth gravitational field strength at distance R from spherical mass M inside a sphere, assuming linearity
Potential energy (Review) The energy transfer when a mass m is moved in a gravitational field PE = mg h Applies when g is considered constant over h - 9.81 Nkg 1 only valid on surface of Earth In order to measure the GPE we need a zero point
Gravitational Potential energy The energy stored as a result of a bodies position in the gravitational field Infinity is defined as reference point where GPE is zero why? This is a reference point which is the same for all gravitational fields The field strength due to all bodies falls at zero at infinity
Gravitational potential V g The GPE per unit mass at a point in the field Gravitational potential energy depends on mass and height Gravitational potential depends on potential energy and mass GPE = mg h V g = GPE m
Calculating gravitational potential Calculate the work that must be done to lift a test mass m from a point in the field to infinity Work is positive GPE of mass must be increased to infinity where GPE is zero Original position must be negative Therefore negative work must be done
F = GMm x 2 To lift the mass from Earth to infinity, a force F equal and opposite to F must be applied The work done by F in moving it a distance x is W = F x = GMm x2 x integrate from to R W = R GMm x 2 x = GMm x = 0 - GMm R R Remember: Work is the change in energy, W = GPE = + GMm R
Since W = GPE and V g = GPE W = + GMm m R GPE = 0 W = GMm R What will be the expression for gravitational potential V? = W m = GMm Rm V g = GM R (Jkg 1 ) Note: = GM R V g falls as 1 R g falls as 1 R 2
Generalize potential V The work necessary per unit mass to take a small mass from the surface of the earth to infinity PE = W = mgh W = gh m But by definition V g = GPE = W m m Hence V g = W m or W = V gm
Important to remember Work done on a system = negative Work done by a system = positive Lifting from surface of Earth to infinity against gravitational field Falling due to gravitational field
Why are V g and GPEnegative? At infinity GPE = 0 Gravity pulls test mass towards Earth Mass will gain KE To conserve energy, PE must decrease It must be negative, since zero was at infinity PE = 0 at
Think about energy conservation If GPE was positive, as it moves towards Earth GPE would increase KE cannot increase, it must decrease You must do work on the system to stop acceleration Work done on the system is negative W = GPE negative
Or think about an energy well
Points of equal height have equal potential Example What is the potential at A? If a body moves from A to B what is the change in V g? How much work must be done in moving a 2 kg mass from A to B?
Gravitational potential difference V g Definition V g is the difference in gravitational potential b/w two points in a gravitational field. It is equal to the work done per unit mass in moving b/w the two points W = m V g
Remember Field lines of gravitational field point in direction in which a small test mass would move when placed in the field Towards the center of earth or down Field is uniform near Earth Field is equally spaced
Example A mass of m = 500. kg is moved from point A, having a gravitational potential of 75.0 J kg -1 to point B, having a gravitational potential of 25.0 J kg -1. (a) What is the potential difference undergone by the mass? (b) What is the work done in moving the mass from A to B? SOLUTION: (a) V g = V B V A = 25.0 75.0 = - 50. J kg -1. (b) W = m V g = 500-50. = - 25000 J. A m Why did the system lose energy during this movement? B
Topic 10: Fields - AHL 10.1 Describing fields Potential difference the gravitational force PRACTICE: A mass m moves upward a without accelerating. (a) What is the change in potential energy of mass-earth system? (b) What is the potential difference undergone the mass? SOLUTION: Use E P = W = Fd cos (a) F = mg and d = h and = 180 so that E P = Fd cos = (mg) h cos 180 = mg h. (b) W = E P so that W = mg h. Thus m V g = W = mg h V g = g h distance h the d by mg
Example A mass m moves upward a distance h without accelerating. (c) What is the gravitational field strength g in terms of V g and h? (d) What is the potential difference experienced by the mass in moving from h = 1.25 m to h = 3.75 m? Use g = 9.81 ms -2. SOLUTION: Use V g = g h. (c) From V g = g h we see that g = V g h. Thus field strength = (d) From V g = g h we see that V g = (9.81)(3.75 1.25) = 24.5 J kg -1. potential difference position change.
Equipotential surfaces Consider the contour map of Mt. Elbert, Colorado. equipotential surface V g = g h which tells us that if h is constant, so is V g. Thus each elevation line clearly represents a plane of constant potential, an equipotential surface.
Imagine Rotation, tilting, and stacking of these equipotential surfaces will produce a 3D image of Mount Elbert: V g V g V g V g If we sketch the gravitational field vectors into our sideways view, we note that the field lines are always perpendicular to the equipotential surfaces.
In reality Of course, on the planetary scale the equipotential surfaces will be spherical, not flat. And the contour lines will look like this: The gravitational field vector g is perpendicular to every point on the equipotential surface V g. g g g
Potential well We know that for a point mass the gravitational field lines point inward. Thus the gravitational field lines are perpendicular to the equipotential surfaces. A 3D image of the same picture looks like this: m
Example Use the 3D view of the equipotential surface to interpret the gravitational potential gradient g = V g r SOLUTION: We can choose any direction for our r value, say the red line: Then g = V g / y. This is just the gradient of the surface. Thus g is the ( ) gradient of the equipotential surface. r V g
W = - V g m mg h = - V g m g = - V h The potential gradient is equal to the negative field strength The negative sign indicates that gravitational field strength points always towards the lower potential GM E R Gradient = - field strength = - V R = - V R Area under graph = work done Force - GMm R 2
Some Mathematics W = F x R GMm x R = - R 2 = GMm R W m = - GM R W m = V V= - GM R V g = - GM R
More mathematical manipulations Since g = - V x At the surface we get g 0 = - V R V = g 0 R - GM E = g R 0R GM E = g 0 R 2 Hence at a distance R from the center of Earth gravitational potential can be written as V = - GM E = - g 0R E = - g R R 0 R E Since at Earth surface R = R E 2
Example Assume Earth is a uniform sphere of radius 6.4 x 10 6 m and mass 6.0 x 10 24 kg. Find a) gravitational potential at i) Earth s surface ii) at a point 6.0 x 10 5 m above surface b) the work done in taking a 5.0kg mass from surface of the Earth to a point 6.0 x 10 5 m above it. c) the work done in taking a 5.0kg mass from the surface of the Earth to a point where earth s gravitational effect is neglible
Solution
Example Potential and potential energy gravitational From V g = E P / m we have E P = m V g. Thus E P = (4)( - 3k - 7k) = 16 kj.
Example Potential and potential energy gravitational