9.2.1Define gravitational potential and gravitational potential energy. Understand that the work done in a gravitational field is independent of path State and apply the expression for gravitational potential due to a point mass State and apply the formula relating gravitational field strength to gravitational potential gradient Determine the potential due to one or more point masses Describe and sketch the pattern of equipotential surfaces due to one and two point masses State the relation between equipotential surfaces and gravitational field lines. Topic 9: Motion in fields 9.2 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 Topic 9: Motion in fields 9.2 Gravitational field,potential,energy F = Gm 1 m 2 /r 2 universal law of gravitation where G = 6.67×10 −11 N m 2 kg −2 E P = -Gm 1 m 2 /r gravitational potential energy where G = 6.67×10 −11 N m 2 kg −2 FYI  The actual proof is beyond the scope of this course.  Note, in particular, the minus sign.

Define gravitational potential and gravitational potential energy. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy E P = -Gm 1 m 2 /r gravitational potential energy where G = 6.67×10 −11 N m 2 kg −2 EXAMPLE: Find the gravitational potential energy stored in the Earth-Moon system. SOLUTION: Use E P = -Gm 1 m 2 /r. E P = -Gm 1 m 2 /r = -(6.67×10 −11 )(7.36×10 22 )(5.98×10 24 )/3.82×10 8 = -7.68×10 28 J. M = 5.98  kg m = 7.36  kg d = 3.82  10 8 m

Define gravitational potential and gravitational potential energy.  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 y 0 as the arbitrary “zero value” of potential energy. The general formula treats r =  as the “zero value”. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy ∆E P = mg∆y local ∆ E P where g = 9.8 m s -2. FYI  The 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 ∆E P = -Gm 1 m 2 (1/r f – 1/r 0 ).

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 d 1 and d 2 :  Along the displacement d 1, gravity does work given by W 1 = mgd 1 cos 180° = -mgd 1.  Along the displacement d 2, gravity does work given by W 2 = mgd 2 cos 90° = 0.  The total work done by gravity is thus W g = -mgd 1. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy W = Fd cos  work where  is the angle between the force F and the displacement d. mgmg d1d1 d2d2

Understand that the work done in a gravitational field is independent of path.  Consider the movement of a ball from the ground to the table via the 6 new displacements s:  Along s 1, s 3 and s 5 gravity does work given by W g = - mgs mgs mgs 5 = -mg(s 1 + s 2 + s 3 ).  Along s 2, s 4 and s 6 gravity does no work. (Why?)  Superimposing d 1 and d 2 from the previous path we see that W g = -mg(s 1 + s 2 + s 3 ) = -mgd 1.  This is exactly the same as we got before! FYI  The work done by gravity is independent of the path followed. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy mgmg s1s1 s2s2 s3s3 s4s4 s5s5 s6s6 d1d1 d2d2

Understand that the work done in a gravitational field is independent of path.  We call any force which does work independent of path a conservative force.  Thus gravity is a conservative force. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy FYI  Only 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. AB Path 1 Path 2

Understand that the work done in a gravitational field is independent of path.  If work is done in a conservative force field then there is an associated potential energy function. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy EXAMPLE: Show that for a conservative force SOLUTION:  From conservation of mechanical energy we have ∆E P + ∆E K = 0 so that ∆E P = -∆E K.  From the work-kinetic energy theorem we have W = ∆E K.  Thus ∆E P = -W = -Fd cos . ∆E P = -W = -Fd cos  potential energy function where F is a conservative force

Understand that the work done in a gravitational field is independent of path.  If work is done in a conservative force field then there is an associated potential energy function. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy EXAMPLE: Show that for the gravitational force F = -mg that the potential energy function is ∆E P = 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 ∆E P = -W = -Fd cos  = -mg∆y cos 180º = mg∆y. ∆E P = -W = -Fd cos  potential energy function where F is a conservative force mg yoyo yfyf d

Define gravitational potential and gravitational potential energy.  We now define gravitational potential as gravitational potential energy per unit mass just as we did in Topic 6: Topic 9: Motion in fields 9.2 Gravitational field,potential,energy E P = -Gm 1 m 2 /r Gravitational potential energy where G = 6.67×10 −11 N m 2 kg −2 ∆V = ∆E P /m Gravitational potential where G = 6.67×10 −11 N m 2 kg −2 V = -Gm/r FYI  The 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. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy ∆V = ∆E P /m Gravitational potential where G = 6.67×10 −11 N m 2 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/r Gm/r 1, m = 5.98×10 24 kg.  At Earth’s surface r 1 = 6.37×10 6 m.  But then r 2 = 5(6.37×10 6 ) = 3.19×10 7 m. Thus ∆V = - Gm(1/r 2 - 1/r 1 ) = - Gm(1/3.19× /6.37×10 6 ) = - Gm( ×10 -7 ) = - (6.67×10 −11 )(5.98×10 24 )( ×10 -7 ) = ×10 7 J kg -1.

Define gravitational potential and gravitational potential energy. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy FYI  A few words clarifying the gravitational potential energy and gravitational potential formulas are in order. E P = -Gm 1 m 2 /rgravitational potential energy V = -Gm/rgravitational potential  Be 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 gravitationalpotential per unit distance. Thus the GPG = ∆V/∆r. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy 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 = ×10 7 J kg -1.  5R E : r 2 = 3.19×10 7 m. Earth: r 1 = 6.37×10 6 m. ∆r = r 2 – r 1 = 3.19× ×10 6 = 2.55×10 7 m.  Thus the GPG = ∆V/∆r = 5.01×10 7 / 2.55×10 7 = 1.96 J kg -1 m -1.

State and apply the formula relating gravitational field strength to gravitational potential gradient.  The gravitational potential gradient is the change in gravitationalpotential per unit distance. Thus the GPG = ∆V/∆r. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy 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] = m 2 s -2.  Then the units of the GPG are [GPG] = [∆V/∆r] = m 2 s -2 /m = m/s 2.

State and apply the formula relating gravitational field strength to gravitational potential gradient. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy EXAMPLE: Show that for the gravitational field near the surface of Earth SOLUTION: Recall that ∆V = ∆E P /m and ∆E P = mg∆y.  Thus ∆V = ∆E P /m ∆V = mg∆y/m ∆V = g∆y g = ∆V/∆y. g = ∆V/∆y local potential gradient FYI  From the last example we already know that the units work out. The above formula only works where g is constant (for small ∆y’s).

State and apply the formula relating gravitational field strength to gravitational potential gradient.  The following general potential gradient (which we will not prove) works over greater range: Topic 9: Motion in fields 9.2 Gravitational field,potential,energy g = -∆V/∆r general potential gradient EXAMPLE: The gravitational potential in the vicinity of a planet changes from ×10 7 J kg -1 to ×10 7 J kg -1 in moving from r = 1.80×10 8 m to r = 2.85×10 8 m. What is the gravitation field strength in that region? SOLUTION: g = -∆V/∆r g = -( × ×10 7 )/(2.85× ×10 8 ) g = / = 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. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy 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 = -(6.67  )(125)/2750 =  J kg -1.  Thus V tot = 4(-3.03  ) =  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. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy 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 = ∆E P /m and the fact that gravitational potential is zero at infinity. ∆E P = m∆V = m(V – V 0 ) = mV = 365(-1.21  ) =  J. r FYI  Since ∆E P = -W we see that the work done in bringing the mass in from infinity is  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: Topic 9: Motion in fields 9.2 Gravitational field,potential,energy m FYI  Generally 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. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy EXAMPLE: Sketch the equipotential surfaces due to two point masses. SOLUTION: Here is the sketch: m m

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. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy m FYI  A 3D image of the same picture looks like this:

State the relation between equipotential surfaces and gravitational field lines. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy 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∆y ∆V∆V

State the relation between equipotential surfaces and gravitational field lines. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy 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. m m

State the relation between equipotential surfaces and gravitational field lines. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy 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. s a d d l e p o i n t

9.2.7Explain the concept of escape speed from a planet Derive an expression for the escape speed of an object from the surface of a planet Solve problems involving gravitational potential energy and gravitational potential. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy

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 v esc 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 v  0. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy M m R r = 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 ∆E K + ∆E P = 0. Then ∆E K + ∆E P = 0 E K – E K0 + E P – E P0 = 0 (1/2)mv 2 – (1/2)mu GMm/r - - GMm/r 0 = 0 (1/2)mv esc 2 = GMm/R Topic 9: Motion in fields 9.2 Gravitational field,potential,energy 0 0 v esc = 2GM/R escape speed PRACTICE: Find the escape speed from Earth. SOLUTION:  M = 5.98  kg. R = 6.37  10 6 m. Thus v esc 2 = 2GM/R = 2(6.67  )(5.98  )/6.37  10 6 v esc = ms -1 (= mph!)

Solve problems involving gravitational potential energy and gravitational potential. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy  The ship MUST slow down and reverse (v becomes -).  The force varies as 1/r 2 so that a is NOT linear.  Recall that a is the slope of the v vs. t graph.

Solve problems involving gravitational potential energy and gravitational potential. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy Escape speed is the minimum speed needed to travel from the surface of a planet to infinity. It has the formula v esc 2 = 2GM/R.

Solve problems involving gravitational potential energy and gravitational potential. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy To escape we need v esc 2 = 2GM/R e. The kinetic energy alone must then be E = (1/2)mv esc 2 = (1/2)m(2GM/R e ) = GMm/R e. This is to say, to escape E = 4GMm/(4R e ). Since we only have E = 3GMm/(4R e ) the probe will not make it into deep space.

Solve problems involving gravitational potential energy and gravitational potential. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy Recall that E p = -GMm/r. Thus ∆E P = -GMm(1/R – 1/R e ).

Solve problems involving gravitational potential energy and gravitational potential. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy The probe is in circular motion so F c = mv 2 /R. But F G = GMm/R 2 = F c. Thus mv 2 /R = GMm/R 2 or mv 2 = GMm/R. Finally E K = (1/2)mv 2 = GMm/(2R).

Solve problems involving gravitational potential energy and gravitational potential. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy The energy given to the probe is stored in potential and kinetic energy. Thus ∆E K + ∆E P = E GMm/(2R) - GMm(1/R–1/R e ) = 3GMm/(4R e ) 1/(2R) - 1/R + 1/R e = 3/(4R e ) 1/(4R e ) = 1/(2R) R = 2R e

Solve problems involving gravitational potential energy and gravitational potential. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy Just know it!

Solve problems involving gravitational potential energy and gravitational potential. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy  From ∆V = ∆E P /m we have ∆E P = m∆V.  Thus ∆E P = (4)( - 3k - - 7k) = 16 kJ

Solve problems involving gravitational potential energy and gravitational potential. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy It is the work done per unit mass in bringing a small mass from infinity to that point.

Solve problems involving gravitational potential energy and gravitational potential. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy V = -GM/r so that V 0 = -GM/R 0. But –g 0 R 0 = -(GM/R 0 2 )R 0 = -GM/R 0 = V 0. Thus V 0 = -g 0 R 0.

Solve problems involving gravitational potential energy and gravitational potential. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy 0.5  10 7 = 5.0  10 6 = R 0. At R 0 = 0.5  10 7 V 0 = -3.8  From previous problem g = -V 0 /R 0 g =  10 7 / 0.5  10 7 = 7.6 m s -2.

Solve problems involving gravitational potential energy and gravitational potential. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy ∆V = ( )  10 7 ∆V = 3.0  10 7 ∆E K = -∆E P E K – E K0 = -∆E P 0 (1/2)mv 2 = ∆E P v 2 = 2∆E P /m v 2 = 2∆V v 2 = 2(3.0  10 7 ) v = 7700 ms -1. This solution assumes probe not in orbit but merely reaches altitude (and returns).

Solve problems involving gravitational potential energy and gravitational potential. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy ∆E P = 3.0  10 7 m. The probe is in circular motion so F c = mv 2 /R. But F G = GMm/R 2 = F c. Thus mv 2 /R = GMm/R 2 or mv 2 = GMm/R. Then E K = (1/2)mv 2 = GMm/(2R). From energy ∆E K = -∆E P or E K – E K0 = -∆E P. Then E K0 = ∆E P + E K = ∆E P + GMm/(2R). From g 0 = GM/R 0 2 we have GM = g 0 R 0 2. Then E K0 = ∆E P + g 0 R 0 2 m/(2R) so that E K0 = m[3  (0.5  10 7 ) 2 /(2  2  10 7 )] = 3.5  10 7 m. Then (1/2)mv 2 = 3.3  10 7 m and v = 8100 ms -1. This solution assumes the satellite is in orbit.

Solve problems involving gravitational potential energy and gravitational potential. Topic 9: Motion in fields 9.2 Gravitational field,potential,energy Just know it!