Gravitational potential Is this the same as gravitational potential energy?

Review

• 𝐹 = −𝐺𝑀𝑚

𝑅2 force b/w two point masses

• g = 𝐹

𝑚 gravitational field strength on Earth

• g = (-)𝐺𝑀

𝑅2 gravitational field strength at distance R

from spherical mass M

• 𝑔𝑥= 4

3𝜋𝐺𝜌𝑥 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 𝑉𝑔

• The GPE per unit mass at a point in the field

GPE = mg∆ℎ 𝑉𝑔= 𝐺𝑃𝐸

𝑚

Gravitational potential

depends on potential energy

and mass

Gravitational potential energy

depends on mass and height

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

𝑥2

• 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 = 𝐺𝑀𝑚

𝑥2∞

𝑅 𝜕𝑥 = −

𝐺𝑀𝑚

𝑥

∞

𝑅= 0 - −

𝐺𝑀𝑚

𝑅= +

𝐺𝑀𝑚

𝑅

Remember: Work is the change in energy, W = ∆ 𝐺𝑃𝐸

• Since W = ∆ 𝐺𝑃𝐸 and

W = + 𝐺𝑀𝑚

𝑅

∆𝐺𝑃𝐸 = 0 −𝑊 = − 𝐺𝑀𝑚

𝑅

• What will be the expression for gravitational potential V?

• 𝑉𝑔= 𝐺𝑃𝐸

𝑚=

𝑊

𝑚= −

𝐺𝑀𝑚

𝑅𝑚 = −

𝐺𝑀

𝑅

𝑉𝑔= −𝐺𝑀

𝑅 (J𝑘𝑔−1)

Note:

𝑉𝑔 falls as 1

𝑅

g falls as 1

𝑅2

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

𝑚 = gh

• But by definition

𝑉𝑔= 𝐺𝑃𝐸

𝑚=

𝑊

𝑚

Hence 𝑉𝑔 = 𝑊

𝑚 or W = ∆𝑉𝑔m

Important to remember

• Work done on a system = negative

• Work done by a system = positive

Falling due to gravitational field Lifting from surface of Earth to infinity against gravitational field

Why are 𝑉𝑔 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 𝑉𝑔?

• How much work must be done in moving a 2 kg mass from A to B?

Gravitational potential difference ∆𝑉𝑔

• Definition ∆𝑉𝑔 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 ∆𝑉𝑔

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

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) Vg = VB – VA = 25.0 – 75.0 = -50. J kg-1.

(b) W = m Vg = 500-50. = -25000 J.

Why did the system lose energy during this movement?

A B m

Example

PRACTICE: A mass m moves upward a distance h without accelerating.

(a) What is the change in potential energy of the mass-Earth system?

(b) What is the potential difference undergone by the mass?

SOLUTION: Use EP = –W = –Fd cos

(a) F = mg and d = h and = 180 so that

EP = –Fd cos = –(mg) h cos 180 = mgh.

(b) W = –EP so that W = –mgh. Thus

m Vg = W = –mgh

Vg = –gh

Potential difference – the gravitational force

Topic 10: Fields - AHL 10.1 – Describing fields

d

mg

A mass m moves upward a distance h without accelerating.

(c) What is the gravitational field strength g in terms of Vg 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.

Example

SOLUTION: Use Vg = – gh.

(c) From Vg = – gh we see that g = – Vg

h . Thus

field strength = – potential difference

position change .

(d) From Vg = – gh we see that

Vg = –(9.81)(3.75 – 1.25) = – 24.5 J kg-1.

Consider the contour map of Mt. Elbert, Colorado.

Vg = – gh which tells us that if h is constant, so is Vg. Thus each elevation line clearly represents a plane of constant potential, an equipotential surface.

Equipotential surfaces

equipotential surface

Rotation, tilting, and stacking of these equipotential surfaces will produce a 3D image of Mount Elbert:

If we sketch the gravitational field vectors into our sideways view, we note that the field lines are always perpendicular to the equipotential surfaces.

Imagine…

g V

g V

g V g

V

In reality…

g g

g

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 Vg.

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:

Potential well

m

Use the 3D view of the equipotential surface to interpret the gravitational potential

gradient g = −∆𝑉𝑔

∆𝑟

SOLUTION: We can choose any direction for our r value, say the red line:

Then g = –∆Vg / ∆y.

This is just the gradient of the surface.

Thus g is the (–) gradient of the equipotential surface.

Example

∆r

∆Vg

• W = - ∆𝑉𝑔m mg∆h = - ∆𝑉𝑔m

• 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 Gradient = - field strength

= - ∆𝑉

𝑅

Area under graph = work done

Force - 𝐺𝑀𝑚

𝑅2

−𝐺𝑀𝐸

𝑅 = -

∆𝑉

𝑅

Some Mathematics

• W = F x R

= - 𝐺𝑀𝑚 𝑥 𝑅

𝑅2

= −𝐺𝑀𝑚

𝑅

𝑊

𝑚 = -

𝐺𝑀

𝑅 𝑊

𝑚= ∆𝑉

∆𝑉= - 𝐺𝑀

𝑅

𝑉𝑔= - 𝐺𝑀

𝑅

More mathematical manipulations

• Since g = - ∆𝑉

𝑥

• At the surface we get

𝑔0= - 𝑉

𝑅

𝑉 = −𝑔0R

• - 𝐺𝑀𝐸

𝑅 = −𝑔0R

𝐺𝑀𝐸 = 𝑔0𝑅2

Hence at a distance R from the center of Earth gravitational potential can be written as

V = - 𝐺𝑀𝐸

𝑅 = -

𝑔0𝑅𝐸2

𝑅 = - 𝑔0𝑅𝐸

Since at Earth surface R = 𝑅𝐸

Example • Assume Earth is a uniform sphere of radius 6.4 x 106m and mass

6.0 x 1024kg. Find a) gravitational potential at i) Earth’s surface ii) at a point 6.0 x 105m above surface b) the work done in taking a 5.0kg mass from surface of the Earth to a point 6.0 x 105m 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

From ∆Vg = ∆EP / m we have ∆EP = m∆Vg.

Thus ∆EP = (4)( -3k – -7k) = 16 kJ.

Example

Potential and potential energy – gravitational

Example

Potential and potential energy – gravitational