Unit 2 Universal Gravitation and Circular Motion

UNIVERSAL GRAVITATION AND CIRCULAR MOTION

Two distinct types of motion have been central to our understanding of the universe.

Terrestrial motion- motion of objects on the Earth.

Celestial motion- motion of objects in the heavens.

Early theories of the universe

As far back as the 6th and 7thcenturies B.C., Greek philosophers proposed that the heavens were composed of eight concentric, transparent spheres.

Each sphere rotated on a different axis and at a different rate, but all were centered about the Earth. This geocentric, or Earth-centered, view of the universe persisted with minor modifications, for over 2000 years.

Nicolaus Copernicus thought that the planetary position could be more easily explained if the sun was the center of the universe. Thus the heliocentric system was born. His theory predicted better the position of the planets than did the geocentric system.

Tycho Brahe (1546-1601) plotted the paths of planets for more than 20 years to an accuracy of 1/1000th of a degree.

Johann Kepler (1571-1630) was a mathematician. He came up with three laws:

1. The planets move about the sun in elliptical paths with the sun at one focus of the ellipse.

2. The straight line joining the sun and a given planet sweeps out equal areas in equal intervals of time.

3. The square of the period of revolution of a planet about the sun is proportional to the cube of its mean distance from the sun. R3/T2 = k = Kepler’s constant.

Newton’s Law of Universal Gravitation

Newton determined that all bodies (masses) anywhere in the universe attract each other. The force of attraction is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.

Formula,

Fg = GmM

R2

Fg = Force of gravity (N)

G = Universal Gravitational Constant

= 6.67x10-11 N.m2/kg2

Mand m= masses in kilograms.

R = distance between the masses from

center to centerin m.

Fg

R

Example

Find the force between two masses of 50 kg and 60 kg separated by a distance of 50 cm.

Fg= GmM

R2

= (6.67x10-11)(50)(60)

0.502

= 8.0x10-7 N

Example

The force of attraction between two masses of 500 kg is

6.0x10-7 N. Find the distance between them.

Fg=GmMR2 = GmM

R2F

R2 = 6.67x10-11(500)(500)

6.0 x 10-7

R = 5.3 m

Example

A person whose mass is 70.0 kg stands on the surface of the

Earth.

(a) What does the person weigh using

the law of universal gravitation?

(b) Use another formula to calculate her

weight.

(a) F = GmMF = 6.67x10-11(70)(5.98 x 1024)

R2(6.37 x 106)2

F = 688 N

(b) F = W = mg = (70)(9.81) = 687 N

F = 687 N

Example

The force between two equal masses is 6.5x10-3 N. The distance between the masses is 1.5 m. Find each mass.

F = GmMm = x

R2M = x

F = Gx2x2 = FR2x = 14,808 kg

R2 G

Answer: 1.5x104 kg

Assignment Text. p.172 #8 - 13

p.694 #4 - 9

Example

What is the gravitational force on a

35.0 kg object standing on the Earth’s

surface?

Example

The gravitational force between two

objects is 3.25x10-6 N. The mass

of one object is twice the mass of the

second object. If the distance between the

masses is 2.00 m, find the mass of each

object.

Ratio method of calculating the

gravitational force

m1 m2 R Fg

Example

The gravitational force between two small

masses A and B when placed at a short distance apart is 3.24x10-7 N. What is the

gravitational force between these objects

if the masses of both A and B are doubled and distance between them is tripled?

Example

The force between two masses is 3.5x10-6 N.

Find the force between them if the masses

are tripled and the distance between them is

halved.

Example

The force of gravity between two

equal masses separated by a distance

of 2.0 m is 5.0x10-7N.

Find the masses.

m=173kg

Cavendish Experiment

In 1798 Cavendish measured the value of

G using a torsion balance.

.

Circular Motion

• Characteristics of circular motion:

1. Object is pulled toward center of circle by a force called CENTRIPETAL (center-seeking) FORCE
2. Acceleration is also directed to center and

is called CENTRIPETAL ACCELERATION.

1. The speed of the object is constant
2. Velocity of the object is constantly changing.
3. Velocity is calculated from the tangent of the circle at any given instant

Fc = mv2ac = v2 (F = ma)

R R

Where:Fc = centripetal force (tension) in N

ac = centripetal acceleration in m/s2

m = mass in kg

R = radius of circle in m

v = speed in m/s

Example

A 2.0 kg rubber stopper travels at 20 m/s in a circular path of radius 10 m. Find the centripetal acceleration and force.

ac = v2Fc = mac

R = (2)(40)

= (20)2

10

ac = 40 m/s2Fc = 80 N

• The time for one revolution is called the PERIOD (T)
• You know that speed = distance/time
• Distance for a circle is the circumference = 2R

v =2R

T

Where:v = speed in m/s

R = radius in m

T = period in s

T = Time for one rotation or revolution

Example

If a car travels in a circular path with a radius of 30 m and it takes 60 s, what is the speed of the car?

R = 30 mv =2R= 2(3.14)(30)

T = 60 s T60

 = 3.14= 3.1 m/s

v = ?

Example

An object with a mass of 2.0 kg travels in a circular path of radius 20 m in 15 s. Find the centripetal force.

m = 2.0 kgFc = mv2v = 2πR = 2π(20) = 8.4 m/s

R = 20 m R T 15

T = 15 s= (2)(8.4)2

Fc = ? 20

Fc = 7.0 N

• Period (T) = time for one revolution
• FREQUENCY (f) = number of revolutions per second

f = 1T = 1

T f

f = frequency r/s or Hz (Hertz)

Example 1

A 10 kg object travels in a circular path at 30 rps. If the radius is 10 m find the centripetal force.

m = 10 kgFc = mv2v = 2πR = 1,903 m/s

R = 10 m R T

f = 30 rps

T = 1/30 = 0.033 s = (10)(1903)2

Fc = ? 10

= 3.6  106 N

Example

What is the centripetal acceleration of a yo-yo being swung in a horizontal circle with a period of 0.20 s? The string is 10 cm long.

T = 0.20 sac = v2v = ?

R = 10 cm = 0.1 m R

ac = ? = 99 m/s2

Example

An 8.0 kg object travels in a circular path at 1,800 rpm. If the radius of the path is 10 m, find

a)the centripetal acceleration

m = 8.0 kg

f = 1,800 rpm = 1,800/60 = 30 rps

T = 1/30 = 0.033 s

R = 10 m

ac = ?ac = v2 v = 2(3.14)(10)

R 0.033

= 3.6  105 m/s2

b) the centripetal force

Fc = mac

= (8.0)(3.6 x 105)

= 2.9  106 N

p. 145 #9-11

p.151 – key terms

p. 166 #5

p. 173 #20 - 24

Banking Curves

• Banking curves often occur so that vehicles do not need to rely only on friction to keep them on the road

tan  = v2

Rg

where: = banking angle

v = speed in m/s

R = radius of circle in m

g = acceleration due to gravity = 9.81 m/s2

Example 1

A car travels around a curve of radius 60.0 m at a speed of 22.0 m/s. At what angle must the curve be banked so the car doesn’t have to rely on friction to stay on the road?

R = 60.0 mtan = v2

v = 22.0 m/s Rg

g = 9.81 m/s2 = (22.0 m/s)2

(60.0 m)(9.81 m/s2)

= 0.8222901801

= 39.43012526

= 39.4

• we have looked at circular motion as horizontal motion
• now lets look at vertical circular motion…we need to incorporate gravity

v = RgEquation #20

where:v = speed in m/s

R = radius in m

g = 9.81 m/s2

Example 1

An amusement park ride spins in a vertical circle. If the diameter of the ride is 5.80 m, what minimum speed must the ride travel at the top?

R = 5.80 m 2v = Rg

= 2.90 m = (2.90 m)(9.81 m/s2)

g = 9.81 m/s2 = 5.3337604 m/s

= 5.33 m/s

Vertical Circular Motion

• now we can look at the forces exerted when an object is being swung in a vertical circle on a

string or rod:

• Fc = centripetal force (net force) and it always acts towards the centre
• FT = tension and it also always acts towards the centre
• Fg = force of gravity and it always acts towards the centre of the Earth

• Fc remains constant (as long as speed remains constant) and Fg remains constant
• FT changes depending on where the mass is in it’s circular path
• tension on the string/rod is greatest at the bottom

Fc(net) = FT – Fg (vectors in opposite direction)

FT = Fc + Fg

• tension on the string/rod is smallest at the top

Fc(net) = FT + Fg (vectors in same direction)

FT = Fc – Fg

Example 1

A 3.0 kg mass moves on a string in a vertical circle of radius 2.5 m with a constant speed of 7.2 m/s.

a) Where is the tension in the string the greatest? Calculate it.

Tension is the greatest at the bottom.

m = 3.0 kgFT = Fc + Fg

r = 2.5 mFT = mv2 + mg

v = 7.2 m/sr

g = 9.81 m/s2 = (3.0 kg)(7.2 m/s)2 + (3.0 kg)(9.81 m/s2)

2.5 m

= 91.638 N

= 92 N

b) Where is the tension in the string the smallest? Calculate it.

Tension is smallest at the top.

m = 3.0 kgFT = Fc – Fg

r = 2.5 mFT = mv2 - mg

v = 7.2 m/sr

g = 9.81 m/s2 = (3.0 kg)(7.2 m/s)2 - (3.0 kg)(9.81 m/s2) 2.5

= 32.778 N

= 33 N

B. Gravitation

• in 1666, Newton used mathematical arguments to look at the force of attraction between two objects
• the falling apple made him realize that the earth was attracting the apple, and that this attraction probably extended far above the earth’s surface into space
• he recognized that the force acting on the apple was proportional to its mass and that, according to his own 3rd law, the apple must also be attracting the earth
• if the earth was involved in the attraction, then the force of attraction was also proportional to the earth’s mass

• it all comes down to this:

1. all objects in the universe attract each other
2. the force of attraction (Fg) between 2 masses (m1, m2)is directly proportional (if one goes up the other goes up) to the product of the masses: Fg m1m2
3. the force of attraction between 2 masses is indirectly proportional (if one goes up the other goes down) to the square of the distance (r) between them: Fg1

r2

• combining these 2 proportionalities gives us:

Fgm1m2

r2

• now, whenever you change from  to =, there is a constant involved
• the formula is:

Fg = Gm1m2Equation #14

r2

where:G = universal gravitation constant = 6.67 x 10-11 Nm2/kg2

m1m2 = mass 1 and mass 2 in kg

Fg = gravitational force of attraction in N

r = distance between centres in m

Example 1

What is the force of attraction between a 20 kg mass and a 50 kg mass separated by a distance of 50 cm?

m1 = 20 kgFg = Gm1m2

m2 = 50 kg r2

r = 50 cm = (6.67x10-11 Nm2/kg2)(20 kg)(50 kg)

= 0.50 m(0.50 m)2

G = 6.67x10-11 Nm2/kg2 = 2.668 x 10-7 N

= 2.7 x 10-7 N

Example 2

A person has a mass of 60 kg.

a) what is his weight on the surface of the earth?

m = 60 kgFg = mg

g = 9.81 m/s2 = (60 kg)(9.81 m/s2)

= 588.6 N

= 5.9 x 102 N

b) what is the gravitational force of attraction between the person and the earth?

m1 = 60 kgFg = Gm1m2

m2 = 5.98x1024 kg r2

r = 6.37 x 106 m = (6.67x10-11 Nm2/kg2)(60 kg)(5.98 x 1024 kg)

G = 6.67x10-11 Nm2/kg2 (6.37 x 106 m)2

= 589.7927146 N

= 5.9 x 102 N

*****NOTE: This is the same number as part a)!!!!!!!!!!!!!!!!!!

Example 3

Two masses, m1 = 5.0 x 106 kg and m2 = 3.0 x 108 kg, exert a force of 5.0 x 10-7 N. Find the distance between them.

m1 = 5.0 x 106 kgFg = Gm1m2

m2 = 3.0 x 108 kg r2

Fg = 5.0 x 10-7 Nr2 = Gm1m2

G = 6.67x10-11 Nm2/kg2 Fg

= (6.67x10-11 Nm2/kg2)(5.0x106 kg)(3.0x108 kg)

5.0 x 10-7 N

= 2.001 x 1011 N

 r = 447325.3849 N

= 4.5 x 105 m

Example 4

Two equal masses are attracted by a gravitational force of 3.8 x 10-8 N. If they are separated by 2.3 m, what is each mass?

m1 =m2Fg = Gm1m2 = Gm2

G = 6.67x10-11 Nm2/kg2 r2 r2

r = 2.3 m m2 = Fgr2

Fg = 3.8 x 10-8 N G

= (3.8 x 10-8 N)(2.3 m)2 (6.67x10-11 Nm2/kg2)

= 3013.793103 kg2

m = 54.89802459 kg

= 55 kg

Example 5

Two masses exert a force of 3.8 x 10-6 N on each other and they are 5.2 m apart. If one mass is 2 times the size of the other, find the mass of each object.

Fg = 3.8 x 10-6 NFg = Gm1m2

r = 5.2 m r2

m1 = m = G(m)(2m)

m2 = 2m r2

G = 6.67x10-11 Nm2/kg2 = 2m2G

r2

m2 = Fgr2

2G

= (3.8 x 10-6 N)(5.2 m)2

2(6.67x10-11 Nm2/kg2)

= 770254.8726 kg2

 m1 = 877.6416538 kg

= 8.8 x 102 kg

 m2 = 2(877.6416538 kg)

= 1755.283308 kg

= 1.8 x 103 kg

Your Assignment: 1. pg 1 #1-10 in workbook

1. pg 172 #8-12

• now we must look at the relationship between force, mass and distance
• if you change one, what happens to the others?

F m1m2

r2

• you can use a chart like this for any formula
• let’s look at circular motion:

Fc = mv2

r

Satellites

• Newton’s “thought experiment”

• Cannon on top of mountain – fire it horizontally
• Cannon is a projectile that falls 4.9 m every second
• If the mountain is high enough, the cannon will fall at the same rate that the earth curves away….orbit!!!!!!
• Mountain would have to be at least 150 km high to be above most of the atmosphere for no air resistance

• Satellites are objects that are projected into space
• They are at a constant height above the planet and are in uniform circular motion
• The force of gravity keeps the satellite in orbit around the earth (or any other planet)

Fg = Fc

• lets look at how to calculate the speed and the periodof an orbiting satellite:

Speed:Fg = Fc

GmsMP = msv2 solve for v

R2 R

GMP = v2

R

v = GMPthis formula is used for R satellites only.

where:v = speed in m/s

G = 6.67 x 10-11Nm2/kg2

MP = mass of planet in kg

R = distance between centre of planet and satellite (radius ofplanet + altitude of the satellite above the surface of theplanet) in m

If orbital radius is given, it is the distance

from the centre of the planet to the satellite.

Example

***A satellite is 100 km above the surface of the earth. What is the speed of the satellite?

G = 6.67 x 10-11Nm2/kg2v = GMP

MP = 5.98 x 1024 kg R

R = 6.37 x 106 m + 1.0 x 105 m

= 6.47 x 106 m

v = ?

= (6.67 x 10-11)(5.98 x 1024)

(6.47 x 106)

v = 7.85 x 103 m/s

Period:

• time taken to complete one revolution.

• some satellites are geosynchronous or geostationary, which means they have a period of 24 hours and they always maintain a specific longitude and latitude
• used for global communications and data collection and transmission eg. weather, minerals, agriculture, astronomy, land and ocean surveillance, solar radiation, the environment etc.
• other satellites are not geosynchronous and you can use a formula to calculate the period (time for one revolution)

v = 2πR

T

T = 2πR

v

Example

*****Find the period of a satellite which is 200 km above the surface of the earth.

G = 6.67 x 10-11Nm2/kg2

MP= 5.98 x 1024 kg

R = 6.37 x 106 m + 2.0 x 105 m

v2 = GMv = 7791 m/s

R

T = 2πRT = 5.3 x 103 s

v

D. Planetary Mechanics

• two different motions have been central to our understanding of the universe:

1.Terrestrial motion – motion of objects on earth

2.Celestial motion – motion of objects in space

• Geocentric view – people used to think that everything in the solar system revolved around the earth; proposed by Greeks
• Heliocentric view – sun-centred motion of planets; proposed by Copernicus (although Galileo was probably the first to propose this, he was thrown in jail)
• Tycho Brahe (1546 – 1601) – Danish astronomer who plotted the paths of planets for more than 20 years to an accuracy of 1/1000 of a degree
• Johannes Kepler (1571 – 1630) – German astronomer who was one of Brahe’s assistants
• he wanted to use a sun-centred system (instead of geocentric) to explain Brahe’s precise data
• he came up with 3 laws:

1.the planets move about the sun is elliptical paths, with the sun at one focus of the ellipse

2.the straight line joining the sun and a given planet sweeps out equal areas in equal time intervals

3.the square of the period of revolution of a planet about the sun is proportional to the cube of its mean distance from the sun

K = R3not on formula sheetbut

T2comes from Equation #12

where:R = mean orbital radius in m

T = orbital period in s

K = Kepler’s constant = 3.315 x 1018 m3/s2

Example 1

An asteroid has a period of 8.1 x 107 s. What is its mean radius of orbit around the sun?

T = 8.1 x 107 sK = R3

K = 3.315 x 1018 m3/s2 T2

R3 = KT2

= (3.315 x 1018 m3/s2)(8.1 x 107 s)2

= 2.1749715 x 1034

R = 2.791372894 x 1011 m

= 2.8 x 1011 m

Example 2

A planets mean distance from the sun is 2.0 x 1011 m. What is the orbital period?

R = 2.0 x 1011 mK = R3

K = 3.315 x 1018 m3/s2 T2

T2 = R3

K

= (2.0 x 1011 m)3

(3.315 x 1018 m3/s2)

= 2.413273002 x 1015 s2

T = 49125075.08 s

= 4.9 x 107 s

Gravitational Field

• Gravitational field surrounds a mass of any size as it is this field that provides the force of attraction on another body
• Knowing the weight (force gravity) and mass of an object, the gravitational field strength can be calculated:

g = Fg

m

where:g = gravitational field in N/kg or m/s2

Fg = gravitational force or weight in N

m = mass in kg

Example 1

On a planet an object weighs 20 N. If the mass of the object is 4.5 kg, what is the gravitational field strength?

Fg = 20 Ng = Fg

m = 4.5 kg m

= 20 N1 N = 1 kg m/s2

4.5 kg

= 4.4 N/kg or m/s2

Example 2

What is the mass of a 40 N object experiencing a gravitational field of 13 N/kg?

Fg = 40 Ng = Fg

g = 13 N/kgm

13 = 40

m

13m=40

m = 3.1 kg

• Gravitational field strength is simply the acceleration due to gravity at the location of the mass e.g. on earth, in orbit, on another planet etc.
• Strength varies inversely with the square of the distance from the center of a planet

g = GM

r2

where:g = gravitational field strength in N/kg or m/s2

G = 6.67 x 10-11 Nm2/kg2

M = mass of planet in kg

r = distance from center of planet in m

Example 1

Find the gravitational field strength on the surface of Mercury (rM = 2.43 x 106 m and M = 3.2 x 1023).

G = 6.67 x 10-11 Nm2/kg2g = GM

r = 2.43 x 106 m r2

M = 3.2 x 1023 kgg = (6.67 x 10-11)(3.2 x 1023)

(2.43 x 106)2

= 3.6 N/kg or m/s2

Example 2

Find the gravitational field strength experienced by astronauts if they are on the space shuttle 500 km above the earth’s surface.

G = 6.67 x 10-11 Nm2/kg2g = GM

M = 5.98 x 1024 kg r2

r = 6.37 x 106 + 500,000 = (6.67 x 10-11)(5.98 x 1024)

(6.87 x 106)2

r = 6.87 x 106 m

g = 8.45 N/kg or m/s2

m (electron) = 9.11 x 10-31 kg

m (proton) = 1.67 x 10-27 kg

F. Einstein’s Theory of Relativity

• use pages 169 – 170 of your text to make short point-form notes on Einstein’s Theory of Relativity

• Einstein proposed that gravity is not a force but an effect of space itself
• He believed that a mass changes the space around it and causes it to be curved
• acceleration takes place because other bodies move in a curved space (centripetal acceleration)
• masses set up a gravitational field around themselves which distorts the space around it

Some Cool Stuff to Think About

• the gravitational field strength of a black hole is so strong that it bends light back into the hole…nothing can leave it
• black holes are formed when a star collapses in on itself
• when the radius is small enough and the collapsing star is dense enough, light can no longer escape
• the earth would have to collapse to a radius of 9 mm (0.009 m), without losing any mass, to become a black hole

so…

*****that’s 50 trillion times higher than the g we experience right now

g = 5.0 x 1013 x 9.81 m/s2

= 4.905 x 1014 m/s2