Gravitation

Forces Causing Centripetal Acceleration

Newton’s Second Law says that the centripetal acceleration is accompanied by a force F = maC = mv2/r F stands for any force that keeps an

object following a circular path Tension in a string Gravity Force of friction

Problem Solving Strategy

Draw a free body diagram, showing and labeling all the forces acting on the object(s)

Choose a coordinate system that has one axis perpendicular to the circular path and the other axis tangent to the circular path

Problem Solving Strategy, cont. Find the net force toward the center of

the circular path (this is the force that causes the centripetal acceleration)

Solve as in Newton’s second law problems: ∑ F = mac (2 dimensions) The directions will be radial and

tangential The acceleration on the right side will be

the centripetal acceleration. Do not put centripetal force on the left

side – it is the right side of the equation already.

Newton’s Law of Universal Gravitation - 1687

“Every particle in the Universe attracts every other particle with a force that is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.”

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r

mmGF

r does not go to zero here because the objects have some radius of their own.

Law of Gravitation, cont.

G is the constant of universal gravitational

G = 6.673 x 10-11 N m² /kg²

This is an example of an inverse square law - force is proportional to 1/r2

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r

mmGF Do the units of G make

sense to you? Could you derive them?

Features of gravitation

1. The gravitational force is a field force that always exists between two particles regardless of the medium that separates them.

2. The force varies as one over the square of the distance between the particles and therefore decreases rapidly with increasing separation.

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Features of gravitation

3. The force is proportional to the product of the particles’ masses.

4. The gravitational force is actually very weak, as shown by the size of G (6.67 x 10-11 Nm2/kg2)

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Gravitation Constant

Determined experimentally by Henry Cavendish in 1798

using this torsion balance. The small spheres m are attracted by large spheres M and the rod rotates. The light beam and mirror serve to amplify the motion.

Quick Quiz – which are false?

Scenario: A ball falls to the ground The force that the ball exerts on the

Earth is equal in magnitude to the force that Earth exerts on the ball.

The ball undergoes the same acceleration as Earth.

Earth is much more massive than the ball, so Earth pulls much harder on the ball than the ball pulls on the Earth. Therefore the ball falls while Earth remains stationary.

Quick Quiz

Superman circles Earth at a radius of 2R where R is Earth’s radius. He then moves out to a radius of 4R. The gravitational force on him at this second orbit as compared to the first is a) the same, b) 2x more, c) 4x more, d) ½ as great, e) ¼ as great.

Example 1 – 2D problem

Find net force on cue ball 1 F21= (6.67E-11)(.3)(.3)/(.4)2

= 3.75E-11 N F31= (6.67E-11)(.3)(.3)/(.3)2

= 6.67E-11 N Fnet = (F21

2 + F312)1/2

=7.65E-11 NCan use ϴ= tan-1(Fy/Fx) to find angle of Fnet

Mass of all balls= 0.3 kg

Example 2– Find total force on B

A CB

4 cm 6 cm

= = 3.33 x 10-6 N

= = 1.78 x 10-6 N

Ftotal = FAB + FBC = -3.33 x10-6 + 1.78 x 10-6 = 1.55 x 10-6 N

mA = 10 kgmB = 8 kgmC = 12 kg

Applications of Universal Gravitation Find the mass of the

earth Use an example of

an object close to the surface of the earth r ~ RE, W = weight

G

gRM

R

mMGmgW

EE

E

E

2

2

Example 2

Find the mass of Earth. Use an object like a baseball of mass m,

falling to Earth. The magnitude of the gravitational force exerted by the Earth on the ball = the weight of the ball.

mg = GMem /RE 2

Cancel m on both sides, solve for Me Me = g RE

2 / G = (9.8) (6.38E6)2/ 6.67E-11 =

5.98E24 kg is the mass of the Earth

Applications of Universal Gravitation

We can rearrange the equation used on the last 2 pages to find the acceleration due to gravity at various radius positions.

g will vary with altitude

2E

r

MGg

(1000 km is about 600 miles)

Gravitational Potential Energy This next section has to do with

Energy. We actually have our Energy unit as Unit 6, later in the semester.

That means you will use some equations here that we haven’t introduced earlier, but trust me, the proof will come later in unit 6.

No matter which order you do Physics in, there are always some before/after entanglements, and this is one of them.

Energy equations Here’s just a quick intro to some of the energy

equations we will be using Potential energy PE = mgh where m=mass,

g=acceleration due to gravity and h = height above the ground (or some reference plane)

Kinetic energy KE = ½ mv2 where m = mass, v = velocity

Work W = Force x distance Work is in Joules, the unit of energy, and it is equal to the net applied force x the distance that the object is moved.

Total energy E = mgh + ½ mv2 This is PE + KE

Gravitational Potential Energy

You may know that potential energy, PEg = mgh This equation is actually only valid when the object

is near Earth’s surface where g is constant. For objects high above Earth’s surface, like a satellite, you need to use this equation, derived by taking

for planetary scale problems Where did the negative sign come from? As two planetary bodies get closer together (r gets

smaller), their PE decreases (so need the negative sign).

Gravitational Potential Energy

Here, PE is zero at infinite distance from Earth’s center, because gravity goes to zero at infinite distance from Earth. (Distant planets can’t feel the pull of the Earth.)

So if r gets smaller, PE decreases! The negative sign indicates that the work done by an

external force in moving an object from infinity to a distance r away from Earth’s center is negative, or the gravitational potential energy decreases in such a process. (it doesn’t gain PE because it gets pulled in w/o an external force)

Another way to say that is that the work is negative because the external force has to hold the object back against the attractive force of gravity.

𝑷𝑬=−𝑮𝑴 𝑬𝒎

𝒓

Potential Energy

Let’s review:• Looking at this graphic,

can you see how the potential energy goes to zero when the object is very far away from Earth?

• Do you see how the PE function is always negative?

• Note that the value of PE at the Earth’s surface is

PE = - GMEm / RE

Total Energy in Planetary Systems

If an isolated system consists of an object of mass m moving with a speed v in the vicinity of a massive object M, the total energy E of the system is the sum of kinetic and gravitational potential energies:

that’s just KE + PE This equation shows that E may be positive,

negative or zero, depending on the value of v. But for a bound system, such as a planet and the Sun, the PE must be < 0 because we have chosen the convention that PE >> 0 as r approaches infinity.

Let’s try setting gravitational force = centripetal force for this orbit.

Total Energy in Planetary Systems (continued) Setting gravitational force = centripetal force

Multiplying both sides by r and dividing by 2 gives

Substituting this back into the energy equation on last page

So total energy is negative and equal in magnitude

to ½ the potential energy.

Escape Speed The escape speed is the speed needed for an object

to soar off into space and not return. We assume that the escape speed is just large

enough to allow the object to reach infinity with a speed of zero. When the object is an infinite distance from Earth its kinetic energy is zero because vf = 0.

The gravitational potential energy is also zero because our zero level of PE was selected at r>>∞, so the sum is zero.

so For the earth, vesc is about 11.2 km/s Note, v is independent of the mass of the object (m)

!

Newton’s Cannon applet and PhET simulation

This applet allows you to see the trajectory of a projectile as a function of launch speed to explore escape speed.

http://waowen.screaming.net/revision/force&motion/ncananim.htm

This PhET simulation models the sun and earth. You can change the masses and watch what happens to the trajectory. You can look it up oneline at PhET too.

gravity-and-orbits_en.jar

Kepler’s Laws – 3 laws

1) All planets move in elliptical orbits with the Sun at one of the focal points.

2) A line drawn from the Sun to any planet sweeps out equal areas in equal time intervals.

3) The square of the orbital period of any planet is proportional to cube of the average distance from the Sun to the planet.

Kepler’s Laws, cont.

Based on observations made by Tyco Brahe

Newton later demonstrated that these laws were the consequence of the gravitational force between any two objects together with Newton’s laws of motion

Kepler’s First Law

All planets move in elliptical orbits with the Sun at one focus.

Any object bound to another by an inverse square law will move in an elliptical path (F ~ 1/r2)

Second focus is empty

Kepler’s Second Law

A line drawn from theSun to any planet will sweep out equal areas in equal times.

Area from A to B and C to D are the same

This is a result of conservation of angular momentum (a topic for AP Physics).

Kepler’s Second Law

Kepler’s Second Law

• So did you notice that when Earth gets closer to the Sun, it has to go faster, to sweep out that area?

• Did you notice that when Earth is farther away from the Sun, it goes slower?

• Why?

If the Earth and the Sun are farther apart, they have more potential energy, because they are attracted to each other.If they are closer together, they have less PE. To make up for losing that PE as Earth approaches the sun, it speeds up, so the total energy will remain the same.

Check out the PhET simulation if you want to see that!

Kepler’s Second Law

• That’s probably very confusing, so let’s just try a numerical example, comparing the PE at 100 units of distance vs. 10 units of distance. Assume that G, ME, m are held constant in each case.

Do you see how PE decreases (gets more negative) when you go from r= 100 to r=10?

Kepler’s Third Law

The square of the orbital period, T of any planet is proportional to cube of the average distance from the Sun to the planet, r.

<< write these eq. on your card Where For orbit around the Sun specifically,

KS = 2.97x10-19 s2/m3

K is independent of the mass of the planet doing the orbiting (it’s a function of the sun only)

Kepler’s Third Law: (T=period)

(TA / TB )2 = (rA/rB)3

Example of using Kepler’s Third If the average distance from the sun for

Venus is 1.082E8 m and for Saturn is 1.434E9 m, and the period of revolution for Saturn is 29.42 Earth years, what is the period for Venus?

let A=Venus, B=Saturn (TA/TB)2=(rA/rB)3 TA = ((rA/rB)3 x TB

2)1/2

TVenus = ((1.082E8/1.434E9)3 x 29.422)1/2

TVenus = 0.61 Earth years

Kepler’s Third Law application

Ms = Mass such as the Sun or other celestial body that has something orbiting it.

Mp = Mass of planet Assuming a circular

orbit is a good approximation

Eccentricity is low for many planets.

Eccentricity, e

Mercury .206 Venus .0068

Earth .0167

Mars .0934

Jupiter .0485

Saturn .0556

Uranus .0472

Neptune .0086

Kepler’s Third Law Derivation

< Set force due to gravity equal to centripetal force

< if orbit is assumed to be circular, then v = distance/time = (2πr/T) where T= period and v = 2πr/T.

< collect r’s all on right, T’s on left, items in parenthesis are constants which are grouped to be called “K”

Example problem (like HW)

What will be the period of a satellite that is orbiting between the orbits of Earth and Mars which has a distance of 1.98 x 1011 m from the sun? Express your answer in years.

= 4.8 x 107 sec

Example problem: Geosynchronous Orbit Satellite dishes do not have to change direction in

order to stay focused on a signal from a satellite. This means that the satellite always has to be found at the same location with respect to the Earth’s surface. For this to occur, the satellite must be at a height such that its revolution period is the same as that of Earth, 24 hours. At what height must a satellite be to achieve this?

The force that produces the centripetal acceleration of the satellite is the gravitational force, so

Ex: Geosynchronous Orbit

Where ME is mass of the Earth, m is mass of the satellite, r is orbital radius from center of Earth.

The velocity of the satellite is . Substituting this in for v in the first equation and solving for r,

this is the radius that will cause a geosynchronous orbit

Professor Lewin – MIT

http://www.youtube.com/watch?v=MJYQGPl3MNI

4’ video on why gravitational force falls away as 1/R^2

Gravity Visualized

https://www.youtube.com/watch?v=MTY1Kje0yLg

9 minute video on visualizing gravity

Ties to PhET simulation “My Solar System”

Space video on what happens when you try to wring water out of a wet wash cloth (great video even if it’s not exactly on the topic of gravity, rather lack of gravity)

http://apod.nasa.gov/apod/ap130424.html