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UniversalGravitation

©2010 by Bryan Pflueger

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Newton's Law of Universal Gravitation

Newton connected the idea that objects, like apples, fall towards the center of Earth with the idea that the moon orbits around the Earth while also falling towards the center of the Earth.

It remains in a circular orbit because it has a velocity perpendicular to its acceleration.

g

v

g

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1 Two objects of mass m are originally separated by distance r and the force acting on each is F. What is the new force if the distance between them is cut in half?

A

B

CDE

rm m

m m

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2 Two object, one of mass m and the other of mass 2m are separated by a distance r. Which if any experiences a greater attractive force?

A mass m

B mass 2m

C Both experience an equal force

D There is no attractive force

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WeightThe weight of an object is defined as the net gravitational force acting upon it. When we are at the surface of a planet, say the Earth, we can ignore the gravitational attraction of the moon.

We also know that the weight can be defined as the mass times the acceleration due to gravity. By using this relation we can find that the acceleration due to gravity is equal to:

(acceleration due to gravity)

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3 An object of mass 10 kg is sitting on the surface of the Earth. What is its weight? (g = 9.8 m/s2)

A 10 N B 49 NC 98 N D 116 N E 980 N

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4 An object of mass 10 kg is sitting on the surface of Mars. What is its mass?

MMars = 6.42x1023 kg RMars = 3.4x106 m

A 10 kgB 19 kg C 38 kg D 76 kg E 98 kg

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5 An object of mass 10 kg is sitting on the surface of Mars. What is its weight?

MMars = 6.42x1023 kg RMars = 3.4x106 m

A 10 N B 19 N C 37 N D 47 N E 76 N

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Work done by Gravity

The work done by the force due to gravity is not dependent on the path that was taken, it is a conservative force which only takes into account the change in position with respect to the center of the planet.

Fg

r1

r2

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6 How much work is required to bring an object of mass m from the surface of the Earth to infinity?

A

B

C

D

E

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7 A rocket ship is sitting on the surface of Neptune. What velocity must it reach in order to leave the planet's surface?

MNeptune = 1.02x1026 kg RNeptune = 2.48x107 m

A 1.66x104 m/s B 2.34x104 m/s C 3.25x104 m/s D 4.5x104 m/s E 7.23x104 m/s

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Circular Orbits

Here is Newton's own drawing of a thought experiment where a cannon on a very high mountain (above the atmosphere) shoots a shell with increasing speed, shown by trajectories for the shell of D, E, F, and G and finally so fast that it never falls to earth, but goes into orbit.

(From M. Fowler, http://galileoandeinstein.physics.virginia.edu/lectures/newton.html)

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Circular OrbitsWhen a satellite, moon, or other object is in an circular orbit about the Earth it is undergoing uniform circular motion. This means that the object is always the same distance from the surface, has a centripetal acceleration (that being gravity), and a velocity perpendicular to its acceleration.

By making these relations we can say that the force due to gravity is equal in magnitude to the centripetal force.

(orbital velocity)

By solving for v, we can find the velocity needed in order to maintain a stable orbit a certain distance r above the planets center.

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8 What velocity must a satellite maintain in order to stay in a circular orbit of one Earth radius above the surface of the Earth?

MEarth = 6x1024 kg REarth = 6.4x106 m

A 2540 m/s B 3700 m/s C 5600 m/s D 6800 m/s E 7900 m/s

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Circular OrbitsBy establishing the relation between the force due to gravity and uniform circular motion we can also solve for the Period, the time it takes to complete one orbit. Period is always represented by T.

By plugging in for v we can solve for the period of the object orbiting a planet.

(Orbital Period)

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9 What is the orbital period for the Earth to orbit the Sun in minutes?

Msun = 2x1030 kg Rorbital = 1.5x1011 m

A 365 mins. B 526729 mins. C 8778 mins. D 31603766 mins. E 54792 mins.

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Circular OrbitsThe equation for the orbital period is given by:

If we rewrite this in a different way we find:

Since the ratio of the square of the period and the cube of the radius give us a constant, from this we can determine the value of another object's orbital period or distance from the planet's center based on an orbit we already know and if given one of the two for the new object.

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10 What is the orbital period for Pluto around the sun given in days?

REarth = 1.5x1011 mRPluto = 5.91x1012 m

A 45134 days B 57082 days C 90268 days D 28750 days E 32000 days

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11 A satellite is in a stable orbit of 3 RE above the surface of the earth. What is the total amount of energy for the satellite?

A

B

C

D

E

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Kepler's Laws1. Each planet moves in an elliptical orbit, with the sun at one focus of the ellipse.

2. A line from the sun to a given planet sweeps out equal areas in equal times

3. The periods of the planets are proportional to the 3/2 powers of the major axis lengths of their orbits.

(elliptical orbit around the sun)

(sector velocity)

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Spherical Mass Distributions By plugging Equation 4 into Equation 3 we find that:

(Eq. 3)(Eq. 4)

To find U we have to write an integral expression.

(For object located outside)

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Spherical Mass Distributions We discussed the case about the potential energy outside of the sphere, but what if you were on the inside? All of the work up to the point where we combine Equations 3 and 4 is exactly the same. The only difference now is a change in the limits for ds.

(For object located inside)

The potential energy is the same anywhere within the sphere.

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Gravitational Force within a Planet

Fg

R

r

In a hypothetical situation, a hole is cut through the Earth and an object is located at a point a distance r away from the planet's center. The force due to gravity will be dependent on the mass within that radius r. For this we exclude every particle of matter beyond r and all the mass located within r is now considered to be concentrated at the center.

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12 A hole is cut straight through a hypothetical planet whose density is uniform throughout. What is the force due to gravity acting on an object of mass m a distance r away from the center of the planet (r is less than the radius of the planet)?

A

B

C

D

E

Fg

R

r

m

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13 A hole is cut straight through a hypothetical planet whose density is uniform throughout. What would be the velocity of an object of mass m at the center of the planet if it were dropped from its surface?

A

B

C

D

E

R

m

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