As a man in his early 20's, Isaac Newton fled Cambridge University to avoid the plague. While living in the country he turned his attention to the problem of planetary motion, hoping to find the underlying cause of the elliptical orbits found by Kepler.

Years later Newton recalled to a friend that he had been sitting in his garden in the afternoon when an apple fell to the ground. It was at this moment that he realized the force which pulled the apple and Earth together was also pulling the moon and Earth together, and the planets and sun together. We accept the idea of gravity so readily today that it's difficult to imagine the importance and profundity of this discovery. In one great generalization Newton solved not only the problem of planetary motion, but the problems of falling objects, buoyancy, the tides, even the missing mass of galaxies that he could not even imagine.

Here is the relationship he discovered:

F = G[(m₁m₂)/r²]

That is, the force acting between two objects is given by the product of their masses divided by the square of the distance between them, multiplied by some number G. The actual value of G, the universal constant of gravitation, was unknown to Newton, but was later determined after his death by Cavendish to be (2/3) * 10^-10 nt-m²/kg².

1. Freida Inhibitions, at 53 kg, sits next to Arthur More, who masses 82 kg. When they are separated by a distance of 1.2 m, what is the force of attraction between them?

F = G[(m1m2)/r2] F = [(2/3) * 10-10 nt-m2/kg2][(53 kg)(82 kg)/(1.22 m 2] F = 2.01 * 10-7 nt note: obviously the relationship is on shaky ground

2. A large mass M is brought close to m on a frictional surface µ. At what distance will m begin to slide towards M?

m slides when Ff = Fg: µmg = G(Mm/r2) r2 = GMm/µmg r = [Gm/µg]1/2

3. At what angle will the attraction of the fixed mass M just keep m from sliding down the incline?

mgsinØ = G(Mm/L2) Ø = sin-1(GM/gL2)

4. A spring of length L and constant k separates two masses, M₁ and M₂. When the masses are released, they will compress the spring slightly. Set up an equation that would give the amount of compression.

kx = G[(m1m2)/(L - x)2] note: this equation requires super algebra to solve, so we will leave it alone!

5. A 50 nt weight hangs vertically at Earth's surface. If a 90 nt weight is brought to a distance of 20 cm, what angle will the string make with the vertical?

tanØ = [((2/3) * 10-10)(50/9.8)(90/9.8)]/(50) tanØ = 1.56 * 10-9 Ø = 8.94 * 10-9°

6. A 4 kg mass rests on a perfectly frictionless surface, 1.7 m away from a 9 kg mass. The two objects are released and their mutual attraction causes them to accelerate.
a) What is the force between them?
b) What is the acceleration of the 4 kg mass?
c) What is the acceleration of the 9 kg mass?

a) F = G[(m1m2)/r2] F = [(2/3) * 10-10][(4 * 9)/(1.72)] F = 8.30 * 10-10 nt

b) F = ma a = F/m a = (8.30 * 10-10)/(4) a = 2.07 * 10-10 m/sec2

c) a = F/m a = (8.30 * 10-10)/(9) a = 9.22 * 10-11 m/sec2 note: mg will slow the ball

7. The mass of Earth is 6 * 10²⁴ kg, and its radius is 6.4 * 10⁶ m. What is the force exerted on a 2 kg mass near its surface?

F = G[(m1m2)/r2] F = [(2/3) * 10-10][(2)(6 * 1024)/(6.4 * 106)2] F = 19.5 nt

8. From the solution of problem 7, find the acceleration of the 2 kg mass if it were released.

a = F/m a = (19.5)/(2) a = 9.75 m/sec2 note: This should look familiar

I hope this looks familiar to you. Notice that the value of the small mass has canceled out, so that all masses will accelerate toward Earth at the same rate. The fact that mass determines the magnitude of the gravitational force as well as the inertia resisting acceleration explains Galileo's discovery that the acceleration of a falling object is independent of its mass.

We can now consider planets in general, and find their surface gravities.

9. A planet of mass M has a radius R. What is the acceleration of a small mass m near its surface?

F = G(Mm/r2) But a = F/m a = [G(Mm/r2)]/m a = GM/r2 note: g is independent of m. Galileo confirmed!

10. The planet Mercury has a mass of 3 * 10²³ kg, and a radius of 2.5 * 10⁶ m. What is the acceleration of gravity at its surface?

From #9: a = GM/r2 a = [(2/3) * 10-10](3 * 1023)/(2.5 * 106)2 a = 3.20 m/sec2 note: thus astronauts on Mercury weigh about 1/3 of their earth weight

11. The sun has a mass of 2 * 10³⁰ kg and a radius of 7 * 10⁸ m. What is the acceleration of gravity near its surface?

a = GM/r2 a = [(2/3) * 10-10](2 * 1030)/(7 * 108)2 a = 272 m/sec2

12. Some stars, in the final stages of their lives, contract to become very small. If a star of mass 5 * 10³⁰ kg collapsed to a radius of 1.2 x 10⁴ m, what would be its surface gravity? How much would a 70 kg baboon weigh standing on it?

a = GM/r2 a = [(2/3) * 10-10](5 * 1030)/(1.2 * 104)2 a = 2.31 * 1012 m/sec2 F = ma F = (70)(2.31 * 1012) F = 1.62 * 1014 nt note: this is about 230,000,000,000 times normal weight! At these forces matter is crushed into a blob of neutrons

Newton looked for confirmation of his law of gravity in the orbit of the moon. He believed that the same force acting on the apple also acted on the moon, and therefore the speed of the moon in its orbit should be predictable by the law of gravity.

13. Given a planet of mass M with a satellite of mass m orbiting at a radius R, what velocity must the satellite have to maintain a circular orbit?

Fc = mv2/r note: this is the force required to hold an object in a circle. Here the force is supplied by gravity GMm/R2 = mv2/r v = (GM/R)1/2

14. Earth has a mass 6 * 10²⁴ kg and a radius 6.4 x 10⁶ m. What is the orbital speed of a satellite that just skims the surface?

v = (GM/R)1/2 v = [((2/3) * 10-10)(6 * 1024)/(6.4 * 106)] v = 7.91 * 103 m/sec

15. Again, given a planet of mass M and satellite of (small) mass m, what is the period of its orbit? (circular orbit of radius R)

v = (GM/R)1/2 T = s/v T = [2(3.14)R]/(GM/R)1/2 T = 2(3.14)R(R/GM)1/2 T = 2(3.14)(R3/GM)1/2

16. The sun has a mass of 2 * 10³⁰ kg. What is the orbital period of Mercury, which is 6 * 10¹⁰ m away?

T = 2(3.14)(R3/GM)1/2 T = 2(3.14)[(6 * 1010)3]/[((2/3) * 10-10)(2 * 1030)] 1/2 T = 8.00 * 106 sec

17. Ganymede, the largest moon of Jupiter, is found to orbit with a period of 7.2 days. It is located 1.07 * 10⁶ km from the center of Jupiter. What is the mass of Jupiter? (Don't forget to convert units to the SI system.)

T = 2(3.14)(R3/GM)1/2 M = [4(3.14)2R3]/[GT2] M = [4(3.14)2(1.07 * 109)3]/[((2/3) * 10-10) (7.2 * 24 * 60 * 60)2] M = 1.87 * 1027

18. At what height would a satellite hover over one spot on Earth?

To stay above one point on a rotating earth: T = (24 hr)(60 min/hr)(60 sec/min) T = 8.64 * 104 sec T = 2(3.14)(R3/GM)1/2 R = [(T2GM)/(4 * (3.14)2)]1/3 R = [[(8.64 * 104)2((2/3) * 10-10)(6 * 1024)]/ [(4(3.14)2)]]1/3 R = 4.23 * 107 m

19. A planet of density r has a satellite that just skims its surface at radius R. What is the period of the satellite?

T = 2(3.14)(R3/GM)1/2 T = 2(3.14)[[R3]/[G(p(4/3)(3.14)R3)]]1/2 T = 2(3.14)[(1)/[Gp(4/3)(3.14)]]1/2 T = [3(3.14)/GP]1/2

20*. Two identical stars each of mass M orbit around their mutual center of mass at a distance R from each other. What is their orbital period? (Be careful, the distance between masses is no longer the orbital radius.)

Fc = Mv2/(R/2) Fg = G(Mm/R2) Set the forces to equal each other: Mv2/(R/2) = G(Mm/R2) v = [(GM)/(2R)]1/2 T = [2(3.14)(R/2)]/v T = (3.14)[(2R3)/(GM)]1/2

21. What is the kinetic energy of a satellite of mass m orbiting a planet of mass M and radius R?

Fg = GMm/R2 Fc = mv2/R Set the forces to equal each other: GMm/R2 = mv2/R But: KE = (1/2)mv2 KE = (1/2)m(Gm/R) KE = GMm/2R

22*. Megan Progress, standing on the surface of a new planet, finds that the local gravity is an astonishing 40 m/sec². She further notes the speed of a satellite just skimming the surface of the planet is 1.2 * 10³ m/sec. What is the radius of this planet?

g = GM/r2 v2 = GM/r To get rid of M, which is not given, solve and substitute: M = v2r/G g = Gm/r2 g = [G(v2r/G)]/r2 g = v2/r r = v2/g r = (1.2 * 103)2/40 r = 3.60 * 104 m

23. Three masses are positioned as shown. What is the resultant force on the 3 kg mass?

The 3 kg mass feels two forces (horizontal and vertical forces) Vertical: Fg = G(Mm/r2) Fg = G[(4 * 1020)(3)]/(2 * 106)2 Fg = .0200 Horizontal: Fg = G(Mm/r2) Fg = G[(5 * 1020)(3)]/(3 * 106)2 Fg = .0111 tanØ = .0200/.0111 Ø = 61.0° F = (.01112 + .02002)1/2 F = .0229 nt

24. Derive the angular momentum of a satellite of mass m orbiting a planet of mass M at a radius R.

v = (GM/R)1/2 L = Iw L = (mR2)(v/R) L = mRv L = mR(GM/R)1/2 L = m(GMR)1/2

When objects lie within a spherical planet they experience a gravity less than they would at the surface!! This is because part of the mass of the planet lies above and is pulling up. In the extreme, at the center of the planet, no force is felt.

To find the mathematical solution for gravity within a planet, we can investigate the effect of a spherical shell of matter on objects within the shell. Place the object, for the purposes of argument, at 1/3 of the way between opposite walls of the sphere. Then matter in the closer wall is only 1/2 the distance away, and thus pulls with 4 times the force. But look at figure 2. We can identify a cone of influence on each side of the mass. Note that within the larger cone there is a circle of matter having twice the radius (and hence 4 times the mass) as there is in the small circle. So.... 1/4 the mass but 4 times the effect works out to the same force on each side of the little mass. Since the shell can be divided similarly into cones that take up the entire shell, there can be no force acting on the mass from the shell.

Now let us return to the depths of an unknown planet of uniform density. One can imagine a planet as being made of concentric spherical shells, something like an onion. The shells that lie outside one's radius have no effect on gravity. Only the mass within one's radius will cause a net effect.

25. A small mass m is located at a distance r from the center of a planet of density p and radius R. What force acts on the mass? (r < R)

Only mass within r contributes to g: F = GMm/r2 F = [G(4/3)(3.14)r3p)m]/r2 F = (4/3)G(3.14)prm

26. Recall that the mass of Earth is 6 * 10²⁴ kg and its radius is 6.4 * 10⁶ m. Assume that the planet has uniform density. What is the acceleration of gravity at a point 2 x 10⁶ m below the surface?

The density of Earth: p = m/V p = (6 * 1024 kg)/[(4/3)(3.14)(6.4 * 106)3 m3] p = 5.47 * 103 kg/m3 g = GM/r2 g = [G(4/3)(3.14)r3p]/r2 g = (4/3)(3.14)Grp g = (4/3)(3.14)G(4.4 * 106)(5.47 * 103) g = 6.71 m/sec2

27. A planet of mass M and radius R has uniform density.
a) What is gravity ";g"; at the surface?
b) What is gravity 1/3 R from the center?

a) g = GM/R2

b) g = G(M'/R'2 note: M' and R' and the reduced mass and radius g = G[(4/3)(3.14)(R/3)3p]/(R/3)2 g = G(4/3)(3.14)(R/3)p g = G(4/3)(3.14)(R/3)[M/[(4/3)(3.14)R3]] g = GM/3R2 note: this is just 1/3 the surface value

Here is a puzzle. If I throw a ball up into the air, it goes to a certain height before returning. If I throw it faster, it goes higher. No matter how hard I throw it, and how high it goes, it will always feel some small force returning it to the planet. Despite this, is it possible to throw a ball so fast that it will never return?

One way to address this problem is to figure out how much work is required to carry a mass out to infinity. If this is a calculable number, then it ought to be theoretically possible to throw a ball with that energy. If it requires an infinite amount of energy, then we won't be able to do it. Recall that F = G(m₁m₂/r²).

As we carry a small mass farther from M, the attractive force on the mass must weaken. In fact it will weaken inversely with the square of the radius. If F_o is the gravitational force on the surface of the planet (R from the center), then a graph of force vs. distance will look like this:

Now think about the definition of work: W = F * s.

The small mass, at any distance from M, will feel some force--call it F. The work required to move the mass a small distance, ÇR, further out must be F * ÇR. To a very good approximation, the force will remain constant over a small ÇR, so we can consider that the total work done is found by adding up the areas of tall, skinny rectangles of height F and width ÇR.

Keep in mind that the work done is the area under the curve. Is the total area out to infinity finite? The non-obvious answer (easily solved by calculus) is that it is finite! In fact, the two areas shown below are equal, and so the energy required to carry m to infinity is just F_o * R = GMm/R² * R = GMm/R.

The units of this may help you to memorize it. GMm/R² has units of force, while R has units of distance. Force times distance, GMm/R² * R = GMm/R has units of work or energy.

29. On a planet of mass M and radius R, a bullet of mass m is fired with an initial velocity vo. How fast is the bullet traveling when it reaches infinity?

KEo - W = KEinfinity (1/2)mvo2 - G(Mm/R) = (1/2)mv2 v = [vo2 - (2GM/R)]1/2

30. On a planet of mass M and radius R, a large spring of constant k is used to launch space vehicles. If the spring is compressed a distance x before firing mass m into deep space, how fast will m be traveling when it reaches infinity?

PEs - Wg = KEinfinity (1/2)kx2 - G(Mm/R) = (1/2)mv2 v = [(kx2/m) - (2GM/R)]1/2

31. An evil alien at infinity bombards Earth with bombs of mass m, fired from a spring released from being compressed an amount x. If the mass and radius of Earth are M and R, with what speed will the bombs strike?

KEs + Wg = KE PEs + Wg = KE (1/2)kx2 + G(Mm/R) = (1/2)mv2 v = [(kx2/m) + (2GM/R)]1/2

By the way, it may sound silly to talk about things at infinity, but to a good approximation, it is not as far away as it sounds. Earth's moon is, in terms of energy, 97% of the way to infinity.

32*. A mass at infinity is thrown with initial velocity vo toward a planet of mass M and radius R. On the way it passes through a cosmic wind tunnel, and then is diverted to skid along the frictional surface of the planet until it stops. How far does it skid?

First calculate g: mg = G(Mm/R2) g = GM/R2 KEo + Wf + Wg = Wf (1/2)mvo2 + FL + G(Mm/R) = µm(GM/R2)s s = [(1/2)mvo2 + FL + G(Mm/R)](R2)/(GµmM)

33. How fast must an object of mass m be ejected from planet M of radius R to just barely reach infinity? That is, minimum initial velocity must it have so as to never return?

KE - Wg = 0 (1/2)mv2 - G(Mm/R) = 0 v = (2GM/R)1/2 note: this is known as the "escape velocity"

This is called the escape velocity from a planet, and is independent of the mass of the object being thrown.

34. What is the escape velocity for an object from Earth?

Ve = (2GM/R)1/2 Ve = [[2((2/3) * 10-10)(6 * 1024)]/(6.4 * 106)] 1/2 Ve = 1.12 * 104 m/sec note: about 25,000 mi/hr

35. A 3 kg meteor in deep space has an initial velocity of 3 * 10⁴ m/sec. How fast will it be traveling when it enters the top cloud layers on Jupiter? (m_j = 2 * 10²⁷kg, r_j = 7 * 10⁷m)

KEinfinity + Wg = KE (1/2)mvo2 + G(Mm/R) = (1/2)mv2 (1/2)(3)(3 * 104)2 + [((2/3) * 10-10)(2 * 1027)(8)]/ (7 * 107 = (1/2)(3)v2 v = 6.86 * 104 m/sec

36. How fast must a projectile be fired from Earth to be traveling at 8000 m/sec when it is at an altitude of 6.4 * 10⁶ m above the surface? (i.e.: one Earth-radius out.)

The work to go from R to infinity is G(Mm/R). The work to go from (R + h) to infinity is G[(Mm)/(R + h)]. So to go from R to R + h requires G(Mm/R) - G[(Mm)/(R + h)] = Wg KEo - Wg = KE (1/2)mvo2 - GMm[(1/R) - (1/(R + h))] = (1/2)mv2 (1/2)vo2 -((2/3) * 10-10)(6 * 1024) [(1/6.4 * 106) - (1/12.8 * 106)] = (1/2)(80002) vo = 1.12 * 104 m/sec

37. The inhabitants of a 3 * 10²⁶ kg planet of 8 * 10⁶ m radius decide to go el-cheapo on launching space probes. A spring whose stiffness is 2 billion nt/m is installed, and masses are fired by releasing the coiled spring. How far would the spring have to be compressed to launch a 5 kg probe to infinity?

PEs = Wg (1/2)kx2 = G(Mm/R) (1/2)(2 * 104)x2 = ((2/3) * 10-10)[(3 * 1026) (5)/(8 * 106)] x = 1.12 * 103

38. Our sun has a mass of 2 * 10³⁰ kg. If it were compressed sufficiently, its surface gravity would be so great light could not escape from it, and it would be called a black hole. What would this new radius be? The speed of light is 3 * 10⁸ m/sec.

Set the escape velocity equal to the speed of light: v = (2GM/R)1/2 (3 * 108) = [(2)((2/3) * 10-10)(2 * 1030)/R]1/2 R = 2.96 * 103

39. What is the escape velocity from a planet of radius R and density p?

(1/2)mv2 = G(Mm/R) v = (2GM/R)1/2 v = [2G((4/3)(3.14)R3p)/R]1/2 v = 2R[(2/3)(3.14)Gp]1/2

For reasons that will become clear after you've worked with the concept a while, the potential energy of a mass at infinity will be considered zero. As the object gets closer to the planet's surface, it has increasingly negative potential energies, given by PE = -GMm/r when it is r from the center. Think of an object as descending into a hole from its initial zero potential energy at infinity.

40. Using the definition given, what is the gravitational potential energy of a 9 kg box of oranges on the surface of Earth? (m_e = 6 * 10²⁴ kg, r_e = 6.4 x 10⁶ m)

PE = -G(Mm/r2) PE = -((2/3) * 10-10)[(6 * 1024)(9)/(6.4 * 106)] PE = -5.62 * 108 J

41. Two stars, each of mass M, are located as shown. What is the potential energy of a mass m when it is located at position A and at position B?

PEA = PE1 + PE2 PE = -G(Mm/R) - G(Mm/2R) PE = -(3/2)G(Mm/R) PEB = PE1 + PE2 PE = -G[(Mm)/(R/2)] - G[(Mm)/(R/2)] PE = -4G(Mm/R)

This concept of negative potential energies is usually troubling to students when they first see it. Think of this analogy. Ground level is assigned a potential energy zero, and two holes of depths h₁ and h₂ are dug. A mass at the bottom of the first hole has a PE₁ = -mgh₁, and in the second hole it will have PE₂ = -mgh₂. The energy required to move the rock from hole 1 to hole 2 is, then:
PE₂ - PE₁ = -mgh₁ -(-mgh₁) = mg(h₁-h₂)

42. A planet of mass M floats around in space.
a) How much energy is required to carry a mass m from r₁ to infinity?
b) ...from r₂ to infinity?
c) ...from r₁ to r₂?

a) Wr1 to infinity = G(Mm/r1)

b) Wr2 to infinity = G(Mm/r2)

c) Wr1 to r2 = Wr1 to infinity - Wr2 to infinity Wr1 to r2 = G(Mm/r1) - GM(m/r2) Wr1 to r2 = GMm[(1/r1) - (1/r2)]

43. A 90 kg meteor falls from infinity toward the sun. If at infinity it had a velocity of 3.0 * 10⁵ m/sec, with what velocity will it enter the sun?

Initial energy = final energy KEo + PEo = KEf + PEf (1/2)(90)(3 * 105) + 0 = (1/2)(90)v2 - G[(2 * 1030)(90)/(7 * 103)] v = 6.86 * 105 m/sec

44. A planet of mass M has a satellite of mass m in a circular orbit of radius R.
a) What is its kinetic energy? b) What is its potential energy? c) What is its total energy?

a) mv2/R = GMm/R2 v = (GM/R)1/2 KE = (1/2)mv2 KE = (1/2)m(GM/R) KE = GMm/2R

b) PE = -G(Mm/R)

c) Etotal = KE + PE Etotal = GMm/2R - GMm/R Etotal = -GMm/2R

45. A rock of mass m is thrown from infinity with an initial velocity v_o. It falls through a cosmic wind tunnel and then compresses a spring on planet M. What is the maximum compression of the spring?

KEo - Ww + Wg = PEs (1/2)mvo2 - FL + G(Mm/R) = (1/2)kx2 x = [[mvo2 - 2FL + 2G(Mm/R)]/(k)]1/2

46. Chuck Roast releases a disk of mass m and radius r from infinity. It has a string wrapped around it so it spins as it falls. At what velocity will it strike a planet of mass M and radius R?

Wg = KEt + KEr G(Mm/R) = (1/2)mv2 + (1/2)Iw2 G(Mm/R) = (1/2)mv2 + (1/2)(1/2 * mr2)(v/r)2 G(Mm/R) = (3/4)mv2 v = [(4GM)/(3R)]1/2