9.2.1Define gravitational potential and gravitational potential energy. Understand that the work done in a gravitational field is independent of path

9.2.1 Define gravitational potential and gravitational potential energy. Understand that the work done in a gravitational field is independent of path.

9.2.2 State and apply the expression for gravitational potential due to a point mass.

9.2.3 State and apply the formula relating gravitational field strength to gravitational potential gradient.

9.2.4 Determine the potential due to one or more point masses.

9.2.5 Describe and sketch the pattern of equipotential surfaces due to one and two point masses.

9.2.6 State the relation between equipotential surfaces and gravitational field lines.

Topic 9: Motion in fields9.2 Gravitational field,potential,energy

Define gravitational potential and gravitational potential energy.

You can think of potential energy as the capacity to do work. And work is a force times a distance.

Recall the gravitational force (from Newton):

If we multiply the above force by a distance r we get


F = Gm1m2/r2 universal law of gravitationwhere G = 6.67×10−11 N m2

kg−2

EP = -Gm1m2/r gravitational potential energy

where G = 6.67×10−11 N m2 kg−2

FYIThe actual proof is beyond the scope of this course.Note, in particular, the minus sign.



EP = -Gm1m2/r gravitational potential energy

where G = 6.67×10−11 N m2 kg−2

EXAMPLE: Find the gravitational potential energy stored in the Earth-Moon system.

SOLUTION: Use EP = -Gm1m2/r.

EP = -Gm1m2/r

= -(6.67×10−11)(7.36×1022)(5.98×1024)/3.82×108

= -7.68×1028 J.

M = 5.981024 kgm = 7.361022

kg

d = 3.82108 m


The previous formula is for large-scale gravitational fields (say, some distance from a planet).

Recall the “local” formula for gravitational potential energy:

The local formula treats y0 as the arbitrary “zero value” of potential energy. The general formula treats r = as the “zero value”.


∆EP = mg∆y local ∆EPwhere g = 9.8 m s-2.

FYIThe local formula works only for g = CONST, which is true as long as ∆y is relatively small (say, sea level to the top of Mt. Everest). For larger distances use ∆EP = -Gm1m2(1/rf – 1/r0).

Understand that the work done in a gravitational field is independent of path.

Recall the definition of work:

Consider the movement of a ball from the ground to the table via displacements d1 and d2:

Along the displacement d1, gravity does work given by W1 = mgd1

cos 180° = -mgd1.

Along the displacement d2, gravity does work given by W2 = mgd2

cos 90° = 0.

The total work done by gravity is thus Wg = -mgd1.


W = Fd cos work where is the angle between the force F and the displacement d.

mg

d1

d2


Consider the movement of a ball from the ground to the table via the 6 new displacements s:

Along s1, s3 and s5 gravity does work given by

Wg = -mgs1 + -mgs3 + -mgs5 = -mg(s1 + s2 + s3).

Along s2, s4 and s6 gravity does no work. (Why?)

Superimposing d1 and d2 from the previous path we see that Wg = -mg(s1 + s2 + s3) = -mgd1.

This is exactly the same as we got before!

FYIThe work done by gravity is independent of the path followed.


mg

s1

s2s3

s4 s5

s6

d1

d2


We call any force which does work independent of path a conservative force.

Thus gravity is a conservative force.


FYIOnly conservative forces have associated potential energy functions.PRACTICE: Show that friction is not a conservative force.SOLUTION: Suppose you slide a crate across the floor along paths 1 and 2:Clearly the distance along Path 2 is greater than on Path 1. The work is different.Thus its work is not path-independent.Thus it is not conservative.

A BPath 1

Pat

h 2


If work is done in a conservative force field then there is an associated potential energy function.


EXAMPLE: Show that for a conservative force

SOLUTION:

From conservation of mechanical energy we have∆EP + ∆EK = 0 so that ∆EP = -∆EK.

From the work-kinetic energy theorem we have W = ∆EK.

Thus ∆EP = -W = -Fd cos .

∆EP = -W = -Fd cos potential energy functionwhere F is a conservative force


If work is done in a conservative force field then there is an associated potential energy function.


EXAMPLE:

Show that for the gravitational force F = -mg that the potential energy function is ∆EP = mg∆y.

SOLUTION:

Consider lifting a weight. The work done by gravity on the weight is independent of the path, so let us lift it straight up for simplicity.

Observe: = 180º, d = ∆y, F = -mg. Thus∆EP = -W = -Fd cos = -mg∆y cos 180º = mg∆y.

∆EP = -W = -Fd cos potential energy functionwhere F is a conservative force

mgyo

yf

d


We now define gravitational potential as gravitational potential energy per unit mass just as we did in Topic 6:


EP = -Gm1m2/r Gravitational potential energy

where G = 6.67×10−11 N m2 kg−2

∆V = ∆EP/m Gravitational potential

where G = 6.67×10−11 N m2 kg−2

V = -Gm/r

FYIThe units of V are J kg-1. Gravitational potential is the work done per unit mass done in moving a small mass from infinity to r. (Note that V = 0 at r = .)

State and apply the expression for gravitational potential due to a point mass.


∆V = ∆EP/m Gravitational potential

where G = 6.67×10−11 N m2 kg−2

V = -Gm/r

EXAMPLE: Find the change in gravitational potential in moving from Earth’s surface to 5 Earth radii (from Earth’s center).

SOLUTION: Use ∆V = -Gm/r2--Gm/r1, m = 5.98×1024 kg.

At Earth’s surface r1 = 6.37×106 m.

But then r2 = 5(6.37×106) = 3.19×107 m. Thus

∆V = -Gm(1/r2 - 1/r1)

= -Gm(1/3.19×107 - 1/6.37×106) = -Gm(-1.26×10-7)

= -(6.67×10−11)(5.98×1024)(-1.26×10-7)

= +5.01×107 J kg-1.



FYIA few words clarifying the gravitational potential energy and gravitational potential formulas are in order. EP = -Gm1m2/r gravitational potential energy V = -Gm/r gravitational potentialBe aware of the difference in name. Both have “gravitational potential” in them and can be confused during problem solving.Be aware of the minus sign on both formulas. The minus sign is there so that as you separate two masses, or move farther out in space, their values increase (as in the last example).Both formulas become zero when r becomes infinitely large.

State and apply the formula relating gravitational field strength to gravitational potential gradient.

The gravitational potential gradient is the change in gravitational potential per unit distance. Thus the GPG = ∆V/∆r.


EXAMPLE: Find the GPG in moving from Earth’s surface to 5 radii from Earth’s center.

SOLUTION: In a previous problem (slide 11) we found the value for the change in gravitational potential to be ∆V = + 5.01×107 J kg-1.

5RE: r2 = 3.19×107

m. Earth: r1 = 6.37×106 m.

∆r = r2 – r1 = 3.19×107 - 6.37×106 = 2.55×107 m.

Thus the GPG = ∆V/∆r

= 5.01×107 / 2.55×107 = 1.96 J kg-1 m-1.


The gravitational potential gradient is the change in gravitational potential per unit distance. Thus the GPG = ∆V/∆r.


PRACTICE: Show that the units for the gravitational potential gradient are the units for acceleration.SOLUTION:The units for ∆V are J kg-1.The units for work are J, but since work is force times distance we have 1 J = 1 N m = 1 kg m s-2 m.Therefore the units of ∆V are (kg m s-2 m)kg-1 or [∆V] = m2 s-2.Then the units of the GPG are [GPG] = [∆V/∆r] = m2 s-2/m = m/s2.



EXAMPLE: Show that for the gravitational field near the surface of Earth

SOLUTION: Recall that ∆V = ∆EP/m and ∆EP = mg∆y.

Thus ∆V = ∆EP/m

∆V = mg∆y/m

∆V = g∆y

g = ∆V/∆y.

g = ∆V/∆y local potential gradient

FYIFrom the last example we already know that the units work out. The above formula only works where g is constant (for small ∆y’s).


The following general potential gradient (which we will not prove) works over greater range:


g = -∆V/∆r general potential gradient

EXAMPLE: The gravitational potential in the vicinity of a planet changes from -6.16×107 J kg-1 to -6.12×107 J kg-1

in moving from r = 1.80×108 m to r = 2.85×108 m. What is the gravitation field strength in that region?

SOLUTION: g = -∆V/∆r g = -(-6.12×107 - -6.16×107)/(2.85×108 - 1.80×108)

g = -4000000/105000000 = -0.0381 m s-2.

Determine the potential due to one or more point masses.

Gravitational potential is derived from gravitational potential energy and is thus a scalar. There is no need to worry about vectors.


EXAMPLE: Find the gravitational potential at the midpoint of the circle of masses shown. Each mass is 125 kg and the radius of the circle is 2750 m.

SOLUTION: Because potential is a scalar, it doesn’t matter how the masses are arranged on the circle. Only the distance matters.

For each mass r = 2750 m. Each mass contributes V = -Gm/r so that

V = -(6.6710-11)(125)/2750 = -3.0310-12 J kg-1.

Thus Vtot = 4(-3.0310-12) = -1.2110-11 J kg-1.

r

Determine the potential due to one or more point masses.

Gravitational potential is derived from gravitational potential energy and is thus a scalar. There is no need to worry about vectors.


EXAMPLE: If a 365-kg mass is brought in from to the center of the circle of masses, how much potential energy will it have lost?

SOLUTION: Use ∆V = ∆EP/m and the fact that gravitational potential is zero at infinity.

∆EP = m∆V

= m(V – V0)

= mV

= 365(-1.2110-11)

= -4.4210-9 J.

r

FYISince ∆EP = -W we see that the work done in bringing the mass in from infinity is +4.4210-9 J.

0

Describe and sketch the pattern of equipotential surfaces due to one and two point masses.

Equipotential surfaces are imaginary surfaces at which the potential is the same.

Since the gravitational potential for a point mass is given by V = -Gm/r it is clear that the equipotential surfaces are at fixed radii and hence are concentric spheres:


m

FYIGenerally equipotential surfaces are drawn so that the ∆Vs for consecutive surfaces are equal.Because V is inversely proportional to r the consecutive rings get farther apart as we get farther from the mass.

equipotential surfaces

Describe and sketch the pattern of equipotential surfaces due to one and two point masses.


EXAMPLE: Sketch the equipotential surfaces due to two point masses.

SOLUTION: Here is the sketch:

mm

State the relation between equipotential surfaces and gravitational field lines.

We know that for a point mass the gravitational field lines point inward.

Thus the gravitational field lines are perpendicular to the equipotential surfaces.


m

FYIA 3D image of the same picture looks like this:



EXAMPLE: Use the 3D view of the equipotential surface to interpret the gravitational potential gradient g = -∆V/∆r.

SOLUTION: We can choose any direction for our r value, say the y-direction:

Then g = -∆V/∆y.This is just the gradient (slope) of the surface.

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

∆y

∆V



EXAMPLE: Sketch the gravitational field lines around two point masses.

SOLUTION: Remember that the gravitational field lines point inward, and that they are perpendicular to the equipotential surfaces.

mm



EXAMPLE: Use a 3D view of the equipotential surface of two point masses to illustrate that the gravitational potential gradient is zero somewhere in between the two masses.

SOLUTION:

Remember that the gravitational potential gradient g = -∆V/∆r is just the slope of the surface.

The slope is zero on the saddle point. Thus g is also zero there.

saddle point

9.2.7 Explain the concept of escape speed from a planet.

9.2.8 Derive an expression for the escape speed of an object from the surface of a planet.

9.2.9 Solve problems involving gravitational potential energy and gravitational potential.


Explain the concept of escape speed from a planet.

We define the escape speed to be the minimum speed an object needs to escape a planet’s gravitational pull.

We can further define escape speed vesc to be that minimum speed which will carry an object to infinity and bring it to rest there.

Thus we see that as r then v0.


Mm

Rr = R r =

Derive an expression for the escape speed of an object from the surface of a planet.

From the conservation of mechanical energy we have ∆EK + ∆EP = 0. Then

∆EK + ∆EP = 0

EK – EK0 + EP – EP0 = 0

(1/2)mv2 – (1/2)mu2 + -GMm/r - -GMm/r0 = 0

(1/2)mvesc2 = GMm/R


0 0

vesc = 2GM/R escape speed

PRACTICE: Find the escape speed from Earth.SOLUTION: M = 5.981024 kg. R = 6.37106 m. Thus vesc

2 = 2GM/R = 2(6.6710-11)(5.981024)/6.37106 vesc = 11200 ms-1 (= 24900 mph!)

Solve problems involving gravitational potential energy and gravitational potential.


The ship MUST slow down and reverse (v becomes -).The force varies as 1/r2 so that a is NOT linear.Recall that a is the slope of the v vs. t graph.



Escape speed is the minimum speed needed to travel from the surface of a planet to infinity.

It has the formula vesc2 = 2GM/R.



To escape we need vesc2 = 2GM/Re.

The kinetic energy alone must then be E = (1/2)mvesc

2 = (1/2)m(2GM/Re) = GMm/Re.This is to say, to escape E = 4GMm/(4Re).

Since we only have E = 3GMm/(4Re) the probe will not make it into deep space.



Recall that Ep = -GMm/r.

Thus ∆EP = -GMm(1/R – 1/Re).



The probe is in circular motion so Fc = mv2/R.

But FG = GMm/R2 = Fc.

Thus mv2/R = GMm/R2 or mv2 = GMm/R.

Finally EK = (1/2)mv2 = GMm/(2R).



The energy given to the probe is stored in potential and kinetic energy. Thus

∆EK + ∆EP = EGMm/(2R) - GMm(1/R–1/Re) = 3GMm/(4Re)

1/(2R) - 1/R + 1/Re = 3/(4Re)1/(4Re) = 1/(2R)

R = 2Re



Just know it!



From ∆V = ∆EP/m we have ∆EP = m∆V.

Thus ∆EP = (4)( -3k - -7k) = 16 kJ



It is the work done per unit mass in bringing a small mass from infinity to that point.



V = -GM/r so that V0 = -GM/R0.

But –g0R0 = -(GM/R02)R0 = -GM/R0 = V0.

Thus V0 = -g0R0.



0.5107 = 5.0106 = R0.

At R0 = 0.5107 V0 = -3.8107.

From previous problem

g = -V0/R0

g = - -3.8107 / 0.5107 = 7.6 m s-2.



∆V = (-0.8 - -3.8)107

∆V = 3.0107

∆EK = -∆EP

EK – EK0 = -∆EP

0

(1/2)mv2 = ∆EP

v2 = 2∆EP/m

v2 = 2∆V

v2 = 2(3.0107)

v = 7700 ms-1.

This solution assumes probe not in orbit but merely reaches altitude (and returns).



∆EP = 3.0107m.The probe is in circular motion so Fc = mv2/R.But FG = GMm/R2 = Fc.Thus mv2/R = GMm/R2 or mv2 = GMm/R.Then EK = (1/2)mv2 = GMm/(2R).From energy ∆EK = -∆EP or EK – EK0 = -∆EP.Then EK0 = ∆EP + EK = ∆EP + GMm/(2R).

From g0 = GM/R02 we have GM = g0R0

2.

Then EK0 = ∆EP + g0R02m/(2R) so that

EK0 = m[3107 + 7.6(0.5107)2/(22107)] = 3.5107m.Then (1/2)mv2 = 3.3107m and v = 8100 ms-1.

This solution assumes the satellite is in orbit.



Just know it!

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9.2.1Define gravitational potential and gravitational potential energy. Understand that the work done in a gravitational field is independent of path