Gravitation & Kepler’s Laws

What causes YOU to be pulled down to the surface of the earth?

THE EARTH….or more specifically…the EARTH’S MASS.

Anything that has MASS has a gravitational pull towards it.

What the proportionality above is

saying is that for there to be a

FORCE DUE TO GRAVITY (Fg) on

something there must be at least 2

masses involved, where one can

be larger than the other.

The force of gravity (Fg) is directly

proportional () to the masses (M

and m) of the objects

𝑭𝒈 M m

As you move AWAY from the earth, your

DISTANCE increases and the FORCE DUE TO

GRAVITY decreases.

This is a special INVERSE relationship called an

Inverse-Square.

The “r” stands for SEPARATION DISTANCE and is the

distance between the CENTERS OF MASS of the 2

objects.

We use the symbol “r” as it often represents the radius.

Gravitation is closely related to circular motion as you

will discover later.

𝑭𝒈𝟏

𝒓𝟐

Fg =m1m2Gr2Force of

attractionSquare of the distancebetween the objects

Masses of two objectsbeing attracted to

each other

Constant of Universal Gravitation,

6.67x1011 m³/kgs²

EXAMPLE:Two planets have the same masses but one is larger than the other. Neither planet has an atmosphere. Which of the following would be the same for objects of equal mass near the surfaces of the two planets?

I. The final velocity of the objects if they are dropped from the same height.II. The momentum of the objects right before hitting the ground if they are

dropped from the same height.III. The centripetal force acting on the masses if they are attached to identical

strings and are spun in vertical circles with the same tangential velocities.

A) I onlyB) II onlyC) I and II onlyD) II and III onlyE) None of the above are the sameE) None of the above are the same

Since the force of gravity is different on each planet because the radii of the planets

are different, the final velocities will be different v = 2gh, the momenta will be

different p = mv, and the centripetal force will be different: Fc = m𝒗𝟐

𝒓.

P.O.D. 1. Two planets have the same size but different masses and no atmospheres. Which of the following would be different for objects of equal mass on the surfaces of the two planets?

I. The rate at which each would fall freely.II. The weight of the objects on weight scales.III. The distance that the objects stretch a spring vertically.

A) I onlyB) II onlyC) I and II onlyD) II and III onlyE) I, II, and III

earth eLEAVING th areyou when thisUse

earth on the areyou when thisUse

2

21

r

mmGF

mgF

g

g

2r

MmGmg

Let’s set the 2 equations for the force

of attraction equal to each other since

they BOTH represent your weight or

force due to gravity

SOLVED FOR g

2

26

2411

s/m81.9)10x37.6(

)10x97.5)(10x67.6(g

Gravity (g) vs. Force of attraction (Fg)?

2r

MGg

mxrkgxM 624 1037.6 Earth theofcenter from dist. ,1097.5Earth theof Mass

ExampleHow far from the earth's surface must an astronaut in space be if she is to feel a gravitational acceleration that is half what she would feel on the earth's surface?

m10x37.6 Earth theof radius r

kg10x97.5Earth theof MassM

6

24

r = 2.64 106 m

This value is four tenths the

radius of Earth.

MG)rr(g)rr(

MGg 2

earth2

earth

g

MG)rr( 2

earth

g

MG)rr( 2

earth

g

MGrr earth

EarthEarth r

g

GMr

62411

10x37.69.4

)10x97.5)(10x67.6(r

2

2

r

MGg

r

MmGmg

P.O.D. 2:A cat weighing 600 N on earth

travels to a planet with 4 times the mass and twice the radius of Earth. The cat's weight on this planet is most nearlyA) 300 N

B) 8002

N

C) 600 N

D) 600 2 NE) 1200 N

Ans. The period of a

pendulum is given by

𝑻 = 𝟐𝝅𝑳

𝒈,

where T is the period (in s)

for the pendulum to go back

and forth, L is the length of

the pendulum (in m), and g

is the value of gravity (in

m/s2)

The Period of a Pendulum

Example(a) What is the period of a pendulum on the

earth’s surface if it has a length of 3 m?

Ans. The period of a pendulum

is given by 𝑻 = 𝟐𝝅𝑳

𝒈

S𝐮𝐛𝐬𝐭𝐢𝐭𝐮𝐭𝐢𝐧𝐠 𝐨𝐮𝐫 𝐠𝐢𝐯𝐞𝐧 𝐯𝐚𝐥𝐮𝐞𝐬:

𝑻 = 𝟐𝝅𝑳

𝒈= 𝟐𝝅

𝟑𝒎

𝟗.𝟖 ൗ𝒎

𝒔𝟐= 𝟑. 𝟒𝟖 𝒔

Example

S𝐮𝐛𝐬𝐭𝐢𝐭𝐮𝐭𝐢𝐧𝐠 𝐨𝐮𝐫 𝐠𝐢𝐯𝐞𝐧 𝐯𝐚𝐥𝐮𝐞𝐬:

𝑻 = 𝟐𝝅𝑳

𝒈= 𝟐𝝅

𝟑𝒎

𝟗.𝟕𝟐 ൗ𝒎

𝒔𝟐= 𝟑. 𝟒𝟗 𝒔

(b) What is the period of the pendulum if it is at a height of 30 km?

First we need to find the value of g at

30 km = 30,000 m

2r

MGg

2

11

r

M)10x67.6(g

m10x37.6 Earth theof radius r

,kg10x97.5Earth theof MassM

6

24

2

26

2411

s/m72.9)1037.6(

)1097.5)(1067.6(g

000,30

Compared to 3.48 s on

the previous slide

P.O.D. 3.A simple pendulum has a period of approximately 2.0 s on earth. What is its period on an imaginary planet the same size as earth, but half the mass of earth?

A) 0.5 sB) 1.0 sC) 1.4 sD) 2.0 sE) 2.8 s

Energy ConsiderationsWork is the integral of a Force function,

F(r), with respect to displacement, r.

REMEMBER: Work was the area under

the curve of the Force vs. distance graph

Substituting in the basic expression for

gravitational force Fg = G𝑀∙𝑚

𝑟2for F(r)

Pulling out the constants and bringing

the denominator to the numerator.

The negative sign should not be surprising as we already knew that Work was

equal to the negative change in “U” or mgh.

Integrating…

This formula gives you the work done by

a mass against the variable force of

gravity as the object gets further and

further away from the planet …

ExampleWhat is the radius of a planet if a rocket

of mass 4000 kg requires 24,000 J of

work to overcome the planet’s force of

gravity? The planet has a mass of

3.6 × 1030 kg.

r

mMGW

W

mMGr

J24000gravityagainstdoneorkwW

kgmN10x67.6 Constant nGravitatio Universal G

,kg 4000 rocket the of Massm

,kg10x6.3Planet the of MassM

2211

30

m002,4J000,24

)kg106.3)(kg4000)(kgNm1067.6(r

302211

P.O.D. 4.A planet has a mass 4 times as great and a radius 2 times as great as that of Planet Earth. How many times more work would a rocket have to do to overcome the pull of gravity on it by the planet?

A) The sameB) Two timesC) Four timesD) Eight timesE) Sixteen times

Consider a rocket leaving the earth. It usually goes up, slows down, and then returns to earth.

There exists an initial minimum speed that when reached the rockets will continue on forever.

Let's use conservation of energy to analyze this situation!...

Escape Velocityspeed needed to escape an object’s

gravitational pull.

Speed greater than vesc.

Speeds smaller

than vesc.

Very far from center of earth: very

small vesc

Farther from center

of earth: small vesc

Surface of earth:

large vesc

Escape Speed

We know that ENERGY will never

change. As the rocket leaves the

earth it's kinetic is large and its

potential is small.

As it ascends, there is a transfer

of energy such that the difference

between the kinetic and potential

will always equal to ZERO.

Escape Speed

Set the potential energy

equal to the kinetic energy

and solve for v, to find the

ESCAPE VELOCITY:

P.O.D. 5.a. How many times greater is the Earth’s escape velocity

compared to Mars’?b. How many times greater is Jupiter’s escape velocity

compared to Earth’s?

Newton’s CannonWhy don’t satellites

fall to the earth ifthey are being pulled

to it by gravity?•If the cannon shoots a cannon ball at a speed

of say 8 m/s, the

following will happen…It falls to earth pulled by

gravity

Newton’s CannonWhy don’t satellites

fall to the earth ifthey are being pulled

to it by gravity?

•If the cannon shootsa cannon ball at say

80 m/s, the

following willhappen…

Newton’s CannonWhy don’t satellites

fall to the earth ifthey are being pulled

to it by gravity?

•If the cannon shootsa cannon ball at slightly less than

8000 m/s, the

following willhappen…The cannon ball will

fall at the base of themountain.

Newton’s Cannon

•If the cannonshoots a cannon

ball at 8000 m/s,

the following willhappen…The cannon ball

will orbit in a circular path

Why don’t satellitesfall to the earth if

they are being pulledto it by gravity?

Newton’s CannonWhy don’t satellites

fall to the earth ifthey are being pulled

to it by gravity?

•If the cannonshoots a cannonball at a velocityslightly over

8000 m/s, what

will happen?The cannon ballwill orbit in anelliptical path

Why doesn’t the earth crash into the sun if it is attracted to it?

Ans: the earth is always “falling” towards the sun. The reason it doesn’t crash into the earth is that it is moving fast

enough that it always “misses”. The earth also has inertia that keeps it on its path.

Kepler's LawsThere are three laws that Johannes Kepler

formulated when he was studying the heavens

THE LAW OF ORBITS - "All planets

move in elliptical orbits, with the Sun at one focus.”

THE LAW OF AREAS - "A line that

connects a planet to the sun sweeps out equal areas in the plane of the planet's orbit in equal times, that is, the rate dA/dt at which it sweeps out area A is constant.”

THE LAW OF PERIODS - "The square

of the period of any planet is proportional to the cube of the semi major axis of its orbit."

Kepler’s 1st law – The Law of Orbits

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

Kepler’s 2nd Law – The Law of Areas

"A line that connects a planet to the sun sweeps out equal areas in the plane of the planet's orbit in equal times, that is, the rate dA/dt at which it sweeps out area A is constant.”

A1 A2

A1 = A2

The planet moves faster near the Sun so the same area is swept out in

a given time as at larger distances, where the planet moves more slowly.

The green arrow represents the planet's velocity, and the purple

arrows represents the force on the planet (the NET force, which is a

combination of the centripetal force and the tangential force)

Kepler’s 3rd Law – The Law of Periods

"The square of the period of any planet is proportional to the cube of the semi major axis of its orbit."

Gravitational forces are centripetal, thus

we can set them equal to each other!

Since we are moving in a circle we can

substitute the appropriate velocity formula!

Using algebra, you can see that everything

in the parenthesis is CONSTANT. Thus the

proportionality holds true!

The expression in the RED circle derived by setting

the centripetal force equal to the gravitational force

is called ORBITAL SPEED.

P. O. D. 6. A satellite is orbiting around the Earth at a radius of 2R from its center. What is the magnitude of the velocity it is traveling at?

A) 𝐺𝑀

𝑅

B) 𝐺𝑀

2𝑅

C) 2𝐺𝑅

𝑀

D) 9.8 m/s

E) 𝐺

𝑅

Kepler’s 3rd Law – The Law of Periods

"The square of the period of any planet is proportional to the cube of the semi major axis of its orbit."

𝑻𝑨𝑻𝑩

𝟐

=𝒓𝑨𝒓𝑩

𝟑

𝒓𝑨

Where TB and rB are the period

and semi major axis of the earth,

which are 1 sidereal year and 1

astronomical unit, the standard.

For planets with CIRCULAR

orbits, the following version of

the formula can also be used:

𝑻𝟐 =𝟒𝝅𝟐

𝑮𝑴 r3

P. O. D. 7.Five planets make circular orbits around a star which is much more massive than any of them. The mass and orbital radius of each satellite is given. Which satellite has the longest period?

A) Mass: 2MRadius: R

B) Mass: M

Radius: 2RC) Mass: M

Radius: RD) Mass: 2M

Radius: RE) Mass: M

Radius: 2R

P.O.D. 6A satellite in orbit around the Earth has a period of one hour. An identical satellite is placed in an orbit having a radius which is nine times larger than the first satellite. What is the period of the second satellite?A) 0.004 hrB) 1/3 hrC) 3 hrD) 9 hrE) 27 hr

Kinetic Energy in Orbit

Using our ORBITAL SPEED

derived from K.T.L and the

formula for kinetic energy

we can define the kinetic

energy of an object in a bit

more detail when it is in

orbit around a body.

The question is WHY? Why do we need a new equation for kinetic

energy? Well, the answer is that greatly simplifies the math. If we use

regular kinetic energy along with potential, we will need both the orbital

velocity AND the orbital radius. In this case, we need only the orbital

radius.

Total Energy of an orbiting body

Notice the lack of

velocities in this

expression as mentioned

in the last slide.

So by inspection we see that the kinetic energy function is always

positive, the potential is negative and the total energy function is negative.

In fact the total energy equation is the negative inverse of the kinetic.

The negative is symbolic because it means that the mass “m” is BOUND

to the mass of “M” and can never escape from it. It is called a BINDING

ENERGY.

Energy from a graphical perspective

As the radius of motion gets

larger. The orbiting body’s

kinetic energy must decrease (

slows down) and its potential

energy must increase ( become

less negative).

By saying become less negative

means that we have defined our

ZERO position for our potential

energy at INFINITY.

Fire

Bac

kwar

ds

Fire

Forwar

ds

50%50%

a) Fire Backwards

b) Fire Forwards

Question: If we have an orbiting Earth satellite and we want to put it in a higher velocity orbit, how can we use the satellite’s thrusters to make the adjustment?

How do you move into a higher velocity orbit?

Backwards = speed up

Forwards = slow down

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30

How do you move into a higher velocity

orbit?1) If you fire backwards thinking you will

speed up the satellite you put it into a larger orbital radius which ultimately SLOWS DOWN the satellite as the KE decreases.

2) By thrusting backwards you are ADDING energy to the system moving the total energy closer to ZERO, this results in a larger radius which also causes the KE to decrease.

3) Fire forwards gently so that you do NEGATIVE WORK. This will cause the satellite to fall into a smaller orbit increasing the KE and increasing the speed. It also makes the potential energy increase negatively because you are moving farther from infinity. As the potential increase the KE again decreases.

8. The Earth exerts the necessary centripetal force on an orbiting satellite to keep it moving in a circle at constant speed. Which of the following statements best explains why the speed of the satellite does not change even though there is a net force exerted on it?A) The satellite is in dynamic equilibrium.B) The acceleration on the satellite is zero.C) The centripetal force is in the direction of the velocity of the

satellite.D) The centripetal force is equivalent to the reaction force caused

from the satellite.E) The centripetal force is perpendicular to the velocity of the

satellite.

10. A satellite orbits the Earth in at a distance of 1 Earth radius from the surface. Its velocity is v. If the satellite moves and enters a new orbit at 3 Earth radii from the Earth's surface, its new velocity will be

A) 𝑣

2

B) 𝑣 3

C)𝑣 2

D) 𝑣

3

E) 𝑣

2

FREE RESPONSE 1. In March 1999 the Mars Global Surveyor (GS) entered its final orbit about Mars, sending data back to Earth. Assume a circular orbit with a period of 1.18 102 minutes = 7.08 103 s and orbital speed of 3.40 103 m/s. The mass of the GS is 930 kg and the radius of Mars is 3.43 106 m.(a) Calculate the radius of the GS orbit.(b) Calculate the mass of Mars.(c) Calculate the total mechanical energy of the GS in this orbit.(d) If the GS was to be placed in a lower circular orbit (closer to the surface of

Mars), would the new orbital period of the GS be greater than or less than the given period?_______Greater than _________Less thanJustify your answer.

(e) In fact, the orbit the GS entered was slightly elliptical with its closest approach to Mars at 3.71 105 m above the surface and its furthest distance at 4.36 105 m above the surface. If the speed of the GS at closest approach is 3.40 103 m s , calculate the speed at the furthest point of the orbit.

FREE RESPONSE 2.A student is given the set of orbital data for some of the moons of Saturn shown below and is asked to use the data to determine the mass MS of Saturn. Assume the orbits of these moons are circular.

(a) Write an algebraic expression for the gravitational force between Saturn and one of its moons.

(b) Use your expression from part (a) and the assumption of circular orbits to derive an equation for the orbital period T of a moon as a function of its orbital radius R.

(c) Which quantities should be graphed to yield a straight line whose slope could be used to determine Saturn’s mass?

(d) Complete the data table by calculating the two quantities to be graphed. Label the top of each column, including units.

(e) Plot the graph. Label the axes with the variables used and appropriate numbers to indicate the scale.

Orbital Period, T(seconds)

Orbital Radius, R(meters)

8.14 104 1.85 108

1.18 105 2.38 108

1.63 105 2.95 108

2.37 105 3.77 108