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
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