GRAVITATIONGRAVITATION

• Prior to Isaac Newton, it was generally believed that gravity (the falling of objects) was something that only occurred on Earth.

• Newton proved that gravity extends indefinitely out into the universe.

• Newton realized that the same force that causes an apple to fall could explain how the Moon orbits the Earth.

• If gravity pulls on the Moon, what holds the Moon up?

• The Moon is moving and wants to travel in a straight line (Newton’s 1st Law).

• What prevents the Moon from flying away?

• The Moon is moving and falling.

The Moon “falls into an orbit”.

• Like a baseball curving downwards as it flies over the field,the combination of its forward velocity and the downward acceleration due to gravity give a curved path.

Newton’s ideas Idea 1The force used to keep an object rotating in a circle depends on the object’s speed and the circle’s

radius in this way:-

F = m v2 / rThis implies that the centripetal acceleration (directed towards the

Centre on the circle) is equal to vv22/r. /r.

Idea 2The Moon is in orbit around the Earth because gravity supplies this centripetal force.

Idea 3

This gravitational force is proportional to 1

(distance from Earth’s centre)2.

There are two places where we can compare the Earth’s gravitational field:

one at the Earth’s surface

and the other at the orbit of

the Moon.

Gravitational accn at the Earth’s surface (ge)

Grav. accn at the distance of the Moon’s orbit (gm)

ge = 1 / (radius of Earth)2

gm 1/ (radius of Moon’s 2

orbit) = (radius of Moon’s orbit) 2

(radius of Earth) 2

2mo

2e

R=

R

Rearranging slightly

to get a numerical value for ge, all we need to do is to insert the centripetal acceleration and the known value of the ratio of the orbital sizes (60/1).

2mo

e m2e

Rg = g

R

Centripetal accn of Moon = v2 / Rmo

First - the Moon’s velocity, v,

= circumference of Moon’s orbit

time for one revolution

= 2πRmo / 2.36 x 106 = 1019m/s

and, second, the accand, second, the accnn of Moon, of Moon,

ggmm = = vv22 = = 1019101922 = = 1.038x101.038x1066

RRmomo R Rmomo RRmomo

= 1.038x10= 1.038x106 6 / (60 x R/ (60 x Ree) )

= 1.038x10= 1.038x1066/(60 x 6.38 x /(60 x 6.38 x 101066))

gm = 0.00271m/s2

Now we can substitute this into our expression for ge ge = Rmo

2 x gm

Re2

where Rmo2 / Re

2 = 602

ggee = 60 = 6022 x 0.00271 m/s x 0.00271 m/s22

ge = 9.8 m/s2

NEWTON’S UNIVERSAL LAW OF GRAVITATION

Between every two objects there is an attractive force, the magnitude of which is directly proportional to the mass of each object and inversely proportional to the square of the distance between the centers of the objects.

1 2

2

F M M

1F

d

G = 6.67 x 10-11 m3/(kg s2)

1 22

1 22

M MF

dM M

F = Gd

Examples: The figure shows an arrangement of

three particles, particle 1 of mass m1 = 6.0 kg and particles 2 and 3 of mass m2 = m3 = 4.0 kgand distance a = 2.0 cm. What is the net gravitational force F1, net on particle 1 due toother particles?

2a

am3

m2

m1

Satellites in Circular Orbits and Apparent

Weightlessness

Satellites in Circular Orbits

• Uniform Circular Motion

• Gravity provides centripetal force

• There is only one speed that a satellite can have if the satellite is to remain in an orbit with a fixed radius.

Why only 1 speed?

• r is measured from the center of the earth

Consequences

• Since 1/r– As r decreases, v increases

• Mass of the satellite is not in the equation, so speed of a massive satellite = the speed of a tiny satellite

Any motion controlled only by gravity is an orbit

Without gravityWith gravity

NEWTON: Gravity explains how planets (andmoons & satellites & etc.) go.

Sun

Several trajectories are possible. . .

Object is effectivelycontinuously fallingtoward the sun . . .. . . But never getsthere!

Circle

F

Imagine launching aball sideways nearEarth . . .

???Two satellites are placed in orbit, one about Mars and the other about Jupiter, such that the orbital speeds are the same. Mars has the smaller mass. Is the radius of the satellite in orbit about Mars less than, greater than, or equal to the radius of the satellite orbiting Jupiter?

Example

• Calculate the speed of a satellite 500 km above the earth’s surface.

• v = 7.30 x 103 m/s16,335 mph

GPS

• Read pages 133 and 134 about the GPS satellites

Satellites about Other Objects

• Satellites are anything that orbit some object such as the Earth, Sun, or other massive object

• When using the equation, replace mE with the mass of central object (i.e. sun)

Example

• Find the mass of a black hole where the matter orbiting it at r = 2.0 x 1020 m move at speed of 7520000 m/s.

• M = 1.70x1044 kg

Find the Period of Orbit

• Period – Time taken for one orbit

• From previous equation• Also v = dist / time

Altitude of TV Satellite

• TV Satellite remain in place relative to the earth– This is called Geosynchronous Orbits

• Find the altitude of a geosynchronous satellite above the earth’s surface.

• Altitude = 3.59x107 m• 22300 mi

Apparent Weightlessness

• Astronauts in the space shuttles and international space station seem to float

• They appear weightless

• They are really falling

Apparent Weightlessness

• If you are in a fast elevator going down,– You feel lighter– You are apparently

lighter

• If you are in a broken elevator going down in free fall– You feel very light– You are apparently

weightless

© 2004 Pearson Education Inc., publishing as Addison-Wesley

Orbital Paths

• Extending Kepler’s Law #1, Newton found that ellipses were not the only orbital paths.

• possible orbital paths– ellipse (bound)

– parabola (unbound)

– hyperbola (unbound)

GRAVITATIONAL POTENTIAL ENERGYis energy an object possesses because of its position in a gravitational field.

The most common use of gravitational potential energy is for an

object near the surface of the Earth where the gravitational acceleration can be assumed to be constant at about 9.8

m/s2. Since the zero of gravitational potential energy can be

chosen at any point (like the choice of the zero of a coordinate system),

the potential energy at a height h above that point is equal to

the work which would be required to lift the object to that height

with no net change in kinetic energy. Since the force required to lift it is equal to its weight, it

follows thatthe gravitational potential energy is equal to its weight

times the height to which it is lifted.

The general expression for gravitational potential energy arises from

the law of gravity and is equal to the work done against gravity to bring a mass to a given point in space. Because of the inverse

square nature of the gravity force, the force approaches zero for large distances, and it makes sense to choose the zero of gravitational potential energy at an infinite distance away. The gravitational potential energy near a planet is then negative, since gravity does positive work as the mass approaches. This negative potential is indicative of a "bound state"; once a mass is near a large body, it is trapped until something can provide enough energy to allow it to escape. The general form of the gravitational potential energy of mass m is:

where G is the gravitation constant, M is the mass of the attracting body, and r is the distance between their centers.

This is the form for the gravitational potential energy which is most

useful for calculating the escape velocity from the earth's gravity. (for multiple particles??)

GMmU

r

• From the work done against the gravity force in bringing a mass in from infinity where the potential energy is assigned the value zero, the expression for gravitational potential energy is

• This expression is useful for the calculation of escape velocity, energy to remove from orbit, etc. However, for objects near the earth the acceleration of gravity g can be considered to be approximately constant and the expression for potential energy relative to the Earth's surface becomes

• where h is the height above the surface and g is the surface value of the acceleration of gravity.

r

GMmU

mghU

• Consider a projectile of mass m, leaving the surface of a planet (or some other astronomical body) with escape velocity v. It has a kinetic energy k given by:

21

2mv

• The projectile also has potential energy U given by:

211

2

m where 6.67 10 is the gravitational constant

kg.s

and is the mass of the planet.

GMmU G

r

M

• When the projectile reaches maximum height v=0

since v(instantaneous velocity) is the derivative of altitude with respect to time. Therefore k = 0. Also as r the potential energy goes to zero. Based on the principal of conservation of energy , the total energy of the projectile at the planet’s surface must also have been zero. Therefore,

21( ) 0

2This yields

GMmrK U mv

2GMv

r