Chapter

Universal Gravitation

Blinn College - Physics 2325 - Terry Honan

. - The Force of Gravity

Introduction

If Isaac Newton had merely written down his three laws of motion he would probably still be known as the most important physicist of all time.  Instead, he went far beyond that; he solved all elementary problems involving mechanics and he found the force law describing gravity.  His theory of gravity is the topic of this chapter.

The history of physics is largely a history of the unification of the fundamental forces.  Before Newton there were separate notions of gravity.  What was known as gravity before Newton was the force holding things to the surface of the Earth; we will call this terrestrial gravity.  It was not clear that this force was the same as what was responsible for the motion of planets, moons and comets; this we will refer to as astronomical gravity.  Newton, with his theory of universal gravitation showed that these two forces were actually the same; he unified terrestrial and astronomical gravity.

Origin of the Inverse Square Law

Newton was trying to understand the force law between point masses (Subscript[m, 1]) and (Subscript[m, 2]) separated by a distance r.  The weight of a body is the gravitational force on it.  Since the weight is proportional to the mass, the force on (Subscript[m, 2]) must be proportional to (Subscript[m, 2]).

F∝(Subscript[m, 2])

Because of Newton's third law the magnitude of the force on (Subscript[m, 1]) is the same. Thus the force on (Subscript[m, 2]) must also be proportional to (Subscript[m, 1]).

F∝(Subscript[m, 1])

How does the force vary with distance?  It should also decrease with the distance r and should go to zero as r approaches infinity. Some possibilities are

F∝(1/r),  F∝(1/(r^2)),  F∝(1/(r^3))  or  F∝((e^(-α r))/(r^2)).

Newton settled on an inverse square law

F∝(1/(r^2)).

He deduced the inverse square law by comparing the ratio of the acceleration of the moon toward the earth to the acceleration due to gravity at the surface of the earth to the ratio of the radius of the moon's orbit to the radius of the earth.  Suppose the force has the form  F∝(r^p) for some power p, where an inverse square law gives p=-2.  Since the acceleration of a body is proportional to the force F∝a we can conclude a∝(r^p).  From this proportionality we can write the ratio

((Subscript[a, 2])/(Subscript[a, 1]))=(((((Subscript[r, 2])/(Subscript[r, 1]))))^p).

We can find the power p using logs.

p=((ln((Subscript[a, 2])/(Subscript[a, 1])))/(ln((Subscript[r, 2])/(Subscript[r, 1]))))

Let (Subscript[a, 2]) and (Subscript[r, 2]) refer to the moon's orbit.  At Newton's time the acceleration of the moon toward the earth could be calculated.  The period of the moon's orbit is (Subscript[T, 2])=27.32 days.  The earth-moon distance, which was measured by parallax, is (Subscript[r, 2])=(Subscript[R, EM])=3.84×(10^8) m.  This gives

(Subscript[a, 2])=(((((2 π)/(Subscript[T, 2]))))^2) (Subscript[r, 2])=(((((2 π)/(27.32×24×3600(QuantityUnit)))))^2)3.84×(10^8)(QuantityUnit)=0.002721 (m/(s^2)).

The acceleration at the surface of the earth is (Subscript[a, 1])=g=9.80 m/(s^2) at a distance given by the earth's radius  (Subscript[r, 1])=(Subscript[R, E])=6.37×(10^6) m.  Combining these expressions gives

p=((ln((Subscript[a, 2])/(Subscript[a, 1])))/(ln((Subscript[r, 2])/(Subscript[r, 1]))))=((ln((0.002721/9.80)))/(ln(((3.84×(10^8))/(6.37×(10^6))))))=-2.00,

which verifies the inverse square law.

Newton realized that there was a crucial flaw in his logic here. He wanted a force law between particles; by a particle it is meant that the distance to an object is large compared to the size of the object. Clearly this is not correct for an object on the surface of the earth. Newton felt that the gravitational force for a spherical body, as long as one is outside of the body, is the same as if all of the mass is at the center. To prove this Newton had to first invent integral calculus and then do a difficult integral. We will refer to this result as the shell theorem and will discuss it shortly.

Newton's Law of Universal Gravitation

Newton's law of gravity is the attractive inverse square law between point masses.  If (Subscript[m, 1]) and (Subscript[m, 2]) are point masses separated by distance r the magnitude of the force between them is

F=G (((Subscript[m, 1])(Subscript[m, 2]))/(r^2)).

This can be written as a vector expression.  Let (Subscript[(OverVector[r]), 1,2]) be the vector from mass 1 to mass 2 and let (Subscript[(OverVector[F]), 12]) be the force on mass 2 due to mass 1.

(Subscript[(OverVector[F]), 12])=-G (((Subscript[m, 1])(Subscript[m, 2]))/(Subscript[r, 1,2]^2)) (Subscript[(OverHat[r]), 1,2])

(Subscript[(OverHat[r]), 1,2]) is the unit vector in the direction of the vector  (Subscript[(OverVector[r]), 1,2]).

(Subscript[(OverHat[r]), 1,2])=((Subscript[(OverVector[r]), 1,2])/(‖ (Subscript[(OverVector[r]), 1,2]) ‖))=((Subscript[(OverVector[r]), 1,2])/(Subscript[r, 1,2]))

The constant G is known as Newton's universal gravitational constant. Gravitational forces between everyday objects are incredibly small. This is reflected in the small value of G in our SI system of units. Newton never knew the value of this constant.  It was eventually measured to be

G=6.673×(10^(-11)) ((N·(m^2))/(kg^2)).

The Force on a Mass Due to a Distribution of Masses

Suppose we want to find the force on a mass m due to a discrete distribution of masses, (Subscript[m, 1]), (Subscript[m, 2]), ... .  Define the vector (Subscript[(OverVector[r]), i]) to be from (Subscript[m, i]) to m. The force of (Subscript[m, i]) to m is

(Subscript[(OverVector[F]), i])=-G ((m (Subscript[m, i]))/(Subscript[r, i]^2))(Subscript[(OverHat[r]), i])

where  (Subscript[(OverHat[r]), i])=(Subscript[(OverVector[r]), i])/(Subscript[r, i]).  The total force is the sum of all these parts  (OverVector[F])=(∑_i) (Subscript[(OverVector[F]), i]). This gives

(OverVector[F])=-G m(∑_i)((Subscript[m, i])/(Subscript[r, i]^2))(Subscript[(OverHat[r]), i]) .

This is just using the idea that force is a vector and that forces add as vectors.

In our discussion of moments of inertia we dealt with continuous distributions. Break up a continuous distribution into an infinite number of infinitesimal pieces; take dm to be one of these infinitesimal pieces. The vector (OverVector[r]) is from dm to m.

(OverVector[F])=-G m∫((OverHat[r])/(r^2))dm .

Example . - Net Force on Moon

(Subscript[M, E])=5.97×(10^24)(QuantityUnit) ,  (Subscript[M, S])=1.99×(10^30)(QuantityUnit) ,  (Subscript[M, M])=7.35×(10^22)(QuantityUnit) ,
(Subscript[R, EM])=3.85×(10^8)(QuantityUnit)  and  (Subscript[R, ES])=1.50×(10^11)(QuantityUnit)

Solution

(Subscript[F, E])=G (((Subscript[M, E])(Subscript[M, M]))/(Subscript[R, EM]^2))=1.975×(10^20)(QuantityUnit) .

(Subscript[F, S])=G (((Subscript[M, S])(Subscript[M, M]))/(Subscript[R, ES]^2))=4.338×(10^20)(QuantityUnit) .

(Subscript[F, net])=(Subscript[F, S])+(Subscript[F, E])=6.31×(10^20)(QuantityUnit)

(Subscript[F, net])=(Subscript[F, S])-(Subscript[F, E])=2.36×(10^20)(QuantityUnit)

(Subscript[F, net])=(Sqrt[(Subscript[F, S]^2)+(Subscript[F, E]^2)])=4.77×(10^20)(QuantityUnit)

The Shell Theorem and Spherical Distributions

Consider a uniform spherical shell of mass M and radius R. If we consider a point mass m a distance r from the center then the shell theorem states that:

for r<R,    (OverVector[F])=(OverVector[0])   and           
for r>R,    (OverVector[F])=-G ((M m)/(r^2))(OverHat[r])

To state this in words: When m is outside the shell (r>R) the shell behaves as if all the mass is at the center. When m is inside the shell (r<R) the total gravitational force on the mass is zero. Clearly, by symmetry, at the exact center of the shell the force should be zero. It is a remarkable property of the inverse square law that it is zero everywhere inside. We will not prove the shell theorem this semester. It will be proven in the second semester course in the analogous case of electrostatics.

As a consequence of the shell theorem we can find the force on a mass m a distance r from the center of any spherical distribution. Break up the sphere into thin concentric shells.

If the shell is inside the radius r the shell behaves as if all  its mass is at the origin.  If it is outside it doesn't contribute to the force.  This  gives the result:

(OverVector[F])=-G ((m (Subscript[M, enclosed]))/(r^2))(OverHat[r]).

where (Subscript[M, enclosed]) is the total mass enclosed by a sphere of radius r.

The Gravitational Field

We will refer to the acceleration due to gravity at some position as the gravitational field (OverVector[g]); typically this no longer will be uniform.  Since the force on a mass m is  (OverVector[F])=m (OverVector[g]) we can define the field at some position by the following procedure.  Add a test mass (Subscript[m, 0]) at some position.  Define the field at that position to be the force divided by the test mass.

(OverVector[g])=((OverVector[F])/(Subscript[m, 0]))

The result is then independent of the test mass.

Now consider a point mass M at the origin. The gravitational field at position (OverVector[r]) is found by placing a test mass (Subscript[m, 0]) there. The force on (Subscript[m, 0]) is

(OverVector[F])=-G (((Subscript[m, 0]) M)/(r^2))(OverHat[r]).

Dividing (Subscript[m, 0]) into this gives

(OverVector[g])=-G (M/(r^2))(OverHat[r])   or   g=‖ (OverVector[g]) ‖=G (M/(r^2)).

This also applies outside of a spherical body. At the surface of a spherical planet of mass M and radius R, the magnitude of the field is

g=G (M/(R^2)).

Example . - Weight on a Different Planet

Solution

g=G (M/(R^2)).

(Subscript[W, 1])=m (Subscript[g, 1])  and  (Subscript[W, 2])=m (Subscript[g, 2])   (Underscript[⟹, Dividing])  ((Subscript[W, 2])/(Subscript[W, 1]))=((Subscript[g, 2])/(Subscript[g, 1]))

(Subscript[g, 1])=G ((Subscript[M, 1])/(Subscript[R, 1]^2))  and  (Subscript[g, 2])=G ((Subscript[M, 2])/(Subscript[R, 2]^2))   (Underscript[⟹, Dividing])  ((Subscript[g, 2])/(Subscript[g, 1]))=(((Subscript[M, 2])/(Subscript[M, 1]))/((((Subscript[R, 2])/(Subscript[R, 1])))^2))

((Subscript[W, 2])/(Subscript[W, 1]))=(((Subscript[M, 2])/(Subscript[M, 1]))/((((Subscript[R, 2])/(Subscript[R, 1])))^2))  ⟹  (W/(Subscript[W, E]))=((M/(Subscript[M, E]))/(((R/(Subscript[R, E])))^2)) =(5/(2^2))

W=(5/(2^2))(Subscript[W, E])=(5/(2^2))×120 lb=150 lb

The Cavendish Experiment

Newton published his theory of universal gravitation in 1687. The value of G was not known at the time; it was not measured until more than a century later in 1798 by Henry Cavendish. Because spherical masses behave the same as point masses, Cavendish could use spherical masses to measure G for point masses. The experiment involved a torsion balance, which consisted of two spherical masses m at the end of a light rigid bar. The bar was suspended from a vertical wire at its center. Under a small torque, the balance would rotate and the amount of rotation was proportional to the torque; τ=κ θ, where the constant of proportionality κ could be measured. Two larger spherical masses M were placed, as shown, and that small gravitational force between m an M caused a slight rotation. The small angle of rotation, gave the torque and that then gave the magnitude of the gravitational force.

Although Cavendish is credited with measuring G for the first time, in reality what he did was find the density and mass of the Earth. Since g=G (Subscript[M, earth])/(Subscript[R, earth]^2), that was equivalent to measuring G. The notation we use in (.) involving the constant G is relatively modern and was not introduced until around 1873.

. - Energy Considerations

Potential Energy of Two Masses

Recall that a conservative force is any force whose work is independent of path.  We can define a potential energy for any conservative force by

Δ U=-W=-∫ (OverVector[F])·d(OverVector[r])

We want to find the potential energy for two masses m and M separated by a distance r. To get this we will first consider M to be at a fixed position, which we will  take as the origin, and move m in its presence along some path and find d U for that path.

Since the force on m due to M is  (OverVector[F])=-G ((M m)/(r^2)) (OverHat[r]),  we get

Δ U=-∫ (OverVector[F])·d(OverVector[r])=-(∫_( (Subscript[(OverVector[r]), i]))^( (Subscript[(OverVector[r]), f])))(-G ((M m)/(r^2)))  (OverHat[r])·d(OverVector[r])

Multiplying a vector by a unit vector gives the component of the vector in the direction of the unit vector.  (OverHat[r])·d(OverVector[r]) becomes the radial component of d(OverVector[r]) which is just dr, the infinitesimal change in the radial variable and note this this is not the same as the magnitude of d(OverVector[r]).

(OverHat[r])·d(OverVector[r])=dr  where  dr=d‖ (OverVector[r]) ‖=‖ (OverVector[r])+d(OverVector[r]) ‖-‖ (OverVector[r]) ‖≠‖ d(OverVector[r]) ‖

With the substitution of dr for (OverHat[r])·d(OverVector[r]), the limits of the integral depend only on the radial distances of the endpoints and it is now a simple single-variable integral.

ΔU=G M m (∫_( (Subscript[r, i]))^( (Subscript[r, f])))((dr)/(r^2))=-G M m ((1/(Subscript[r, f]))-(1/(Subscript[r, i])))

Clearly this result is independent of the path taken; this demonstrates the force is conservative and justifies the definition of the potential energy.

We are looking for some function U(r), describing the potential energy as a function of position that satisfies Δ U=U((Subscript[r, f]))-U((Subscript[r, i])). This is unique up to an arbitrary constant. The simplest choice is

U(r)=-G((M m)/r)

where we chose to make the potential energy zero at infinity.

U(∞)=0   or more precisely   (lim_(r->∞))U(r)=0

Potential Energy of a Configuration

Now we consider the case of several point masses, (Subscript[m, 1]), (Subscript[m, 2]), ... .  Take the distance between (Subscript[m, 1]) and (Subscript[m, 2]) to be (Subscript[r, 1,2]) and generally (Subscript[r, i,j]) as the distance between the (i^th) and (j^th) masses.  For two masses we have U=-G (Subscript[m, 1]) (Subscript[m, 2])/(Subscript[r, 1,2]).  For three masses  we have this term plus energy terms between (Subscript[m, 1]) and (Subscript[m, 3]), and between (Subscript[m, 2]) and (Subscript[m, 3]).  We get

U=-G (((Subscript[m, 1]) (Subscript[m, 2]))/(Subscript[r, 1,2]))-G (((Subscript[m, 1]) (Subscript[m, 3]))/(Subscript[r, 1,3]))-G (((Subscript[m, 2]) (Subscript[m, 3]))/(Subscript[r, 2,3])).

The general expression is

U=-G (∑_(i<j)) (((Subscript[m, i]) (Subscript[m, j]))/(Subscript[r, i,j])).

The sum is over all indices i and j where we insist that i<j to avoid double counting.

Energy and Escape Speed

The total energy is the sum of the kinetic energies of all masses and the total potential energy above.  For a single mass m moving in  the presence of a large mass M the total mechanical energy is

E=K+U=(1/2)m (v^2)-G ((M m)/r).

The potential energy is negative, going to zero at infinity.  To escape the gravitational pull of M, m must have enough energy to reach infinity, or zero energy.

The escape speed is the critical speed needed to escape the gravity of a planet from the surface of the planet of mass M and radius R.  This becomes

0=E=(1/2)m (Subscript[v, esc]^2)-G ((M m)/R).

Solving for the escape speed we get

(Subscript[v, esc])=(Sqrt[((2 G M)/R)]).

Example . - Rocket Launched off a Spherical Planet

Solution

E=(1/2)m (v^2)-G ((M m)/r)

(Subscript[E, i])=(Subscript[E, f])  ⟹  (1/2)m (Subscript[v, 0]^2)-G ((M m)/R)=0-G ((M m)/(Subscript[r, max]))  ⟹  (Subscript[r, max])=((((1/R)-((Subscript[v, 0]^2)/(2 G M))))^(-1))

Solution

(Subscript[E, i])=(Subscript[E, f])  ⟹  (1/2)m (Subscript[v, 0]^2)-G ((M m)/R)=(1/2)m (Subscript[v, ∞]^2)+0  ⟹  (Subscript[v, ∞])=(Sqrt[(Subscript[v, 0]^2)-2G (M/R)])

. - Orbits

Circular Orbits

We will now consider circular orbits of a satellite of mass m orbiting a much larger object of mass M,  m≪M.  Take the larger object to be a sphere of radius R. If the satellite orbits at a height h above the surface, then the radius of the orbit is

r=R+h.

We now apply the second law to the satellite.  The only force is gravity and the acceleration is centripetal.  We get:

(Subscript[F, net,c])=m (Subscript[a, c])  ⟹  G ((M m)/(r^2))=m (Subscript[a, c]).

In this expression the satellite mass cancels; this is the essence of weightlessness.  Suppose one studies the orbit of the international space station, then m is the station's mass. If an astronaut is floating in the station and not touching anything, then m is the astronaut's mass. The cancellation of m implies that the astronaut and station have the same orbit; he will float relative to the station.

After canceling the masses we can then use the two expressions for the centripetal acceleration for uniform circular motion.  This gives a pair of expressions, one that relates the speed and radius and the other relating the period and radius.

G (M/(r^2))=(Subscript[a, c])=(Piecewise[{{((v^2)/r)           }, {(((((2 π)/T)))^2) r}}])

Solving these expressions for the speed and period gives.

v=(Sqrt[((G M)/r)])

(T^2)=((4 (π^2))/(G M)) (r^3)

Example . - Low-earth Orbit

Solution

(Subscript[R, earth])=6.38×(10^6)(QuantityUnit) ,  (Subscript[M, earth])=5.97×(10^24)(QuantityUnit)

h=200(QuantityUnit)=2.0×(10^5)(QuantityUnit)   ⟹  r=(Subscript[R, earth])+h=6.58×(10^6)(QuantityUnit)

(T^2)=((4 (π^2))/(G (Subscript[M, earth]))) (r^3)  ⟹  T=5310(QuantityUnit)=89(QuantityUnit)

v=(Sqrt[((G (Subscript[M, earth]))/r)])=7780(QuantityUnit)/(QuantityUnit)

Example . - Geostationary Orbit

Solution

(T^2)=((4 (π^2))/(G M)) (r^3)

(Subscript[T, 1]^2)=((4 (π^2))/(G (Subscript[M, earth]))) (Subscript[r, 1]^3)  and  (Subscript[T, 2]^2)=((4 (π^2))/(G (Subscript[M, earth]))) (Subscript[r, 2]^3)   (Underscript[⟹, Dividing])  (((((Subscript[T, 2])/(Subscript[T, 1]))))^2)=(((((Subscript[r, 2])/(Subscript[r, 1]))))^3)

(Subscript[r, 2])=(((((Subscript[T, 2])/(Subscript[T, 1]))))^(2/3))(Subscript[r, 1])≃(((((24(QuantityUnit))/(1.5(QuantityUnit)))))^(2/3))(Subscript[R, E])=(16^(2/3))(Subscript[R, E])  ⟹  (Subscript[r, 2])≃6.4 (Subscript[R, E])

Solution

(T^2)=((4 (π^2))/(G (Subscript[M, earth]))) (r^3)  ⟹  r=(((G (Subscript[M, earth])((T^2)/(4 (π^2)))))^(1/3))

r=((((6.673×(10^(-11)) ((N·(m^2))/(kg^2))) (5.97×(10^24)(QuantityUnit)) ((((23.93×3600(QuantityUnit)))^2)/(4 (π^2)))))^(1/3))=4.22×(10^7)(QuantityUnit)

(Subscript[R, earth])=6.38×(10^6)(QuantityUnit)   ⟹  r=6.61 (Subscript[R, earth])

Kepler's Laws

Kepler came after both Copernicus and Galileo, and came before Newton. Kepler used data taken by someone else, Tycho Brahe, of the positions of the planets in the sky as functions of time. From this he realized that the circular orbits of Copernicus and Galileo could not fit the data. He settled on elliptical orbits and summarized his results with three laws.

After Kepler, Newton found his three laws of motion and his law of universal gravitation. He was able to derive Kepler's three law from his work. This was a remarkable feat. He was also able to prove that Kepler's laws implied that the gravitational force law had to be an inverse square law.

Kepler's First Law  -  Planets move in elliptical orbits about the sun, which is at a focal point of the ellipse.

Given two points, called focal points, an ellipse is the set of points in a plane such that the sum of the two distances from each focus is a constant.

a is called the semimajor axis; it is half the largest distance between two points on the ellipse.  The other dimension of the ellipse is the minor axis 2 b.  Each focus is a distance of c from the center, where c=e a, with e known as the eccentricity.  The eccentricity is the deviation from being a circle, 0≤e<1.  e=0 is the circular case, where the two foci merge and a=b. As e approaches one, the ellipse approaches a parabola. Proving Kepler’s first law is not that difficult, but is a bit beyond the traditional scope of this course.

Kepler's Second Law  -  The radial vector from the sun to the planet sweeps out equal areas in equal time.

Kepler’s second law describes how fast a planet moves in its orbit; when near the sun it moves rapidly and and far away it moves slowly. It sweeps out equal areas in equal time. This is a simple consequence of the conservation of angular momentum.

The area of a parallelogram formed by two vectors is A B sin θ which is just the magnitude of the cross product of the vectors. In a time dt a planet moves by (OverVector[v]) dt. This traces out a triangle; this is half a parallelogram and thus has the infinitesimal area

dA=(1/2)‖ (OverVector[r])×(OverVector[v]) dt ‖

Since (OverVector[L])=(OverVector[r])×(OverVector[p]) is constant, it follows that dA/dt is constant.

((dA)/(dt))=(1/2)‖ (OverVector[r])×(OverVector[v]) ‖=(1/(2 m))‖ (OverVector[r])×(OverVector[p]) ‖=(L/(2 m))=constant.

This proves the result.

Kepler's Third Law  -  (Period^2)∝(((Semimajor axis))^3)   or    (T^2)∝(a^3)

Define the major axis as the largest distance between two points on an elliptical orbit.  The semimajor axis, a, is half this.  For a circular orbit, this is the same as the radius a=r.  We saw that (T^2)∝(r^3) for a circular orbit.  The third law is then a generalization of this result, which we will not prove.

(T^2)=((4 (π^2))/(G M)) (a^3).