Copyright Sautter 2003

Gravitation

• The Law of Universal Gravitation is based on the observed fact that all masses attract all other masses. The force of attraction decreases as the distance between the masses increases.

• This relationship is called an inverse square law since the decrease in attraction between objects is relative to the square of that distance.

• If the distance between the masses doubles, the force of gravitational attraction becomes ¼ of the original force. If the distance triples, the force becomes 1/9 of the original, as so on.

• For example, 22 = 4, 32 = 9, etc.

Weight & Mass

• Weight and mass measure different things. • Mass measures the quantity of matter which is present. It

represents the inertia property of matter meaning its ability of resist changes in motion. Mass is measured in grams, kilograms or slugs.

• Weight is a force resulting from the effect of gravity on a mass. A mass without gravity is weightless. Weight is measured in dynes, newtons or pounds.

• Gravity on the Earth’s surface is measured as – 980 cm/s2, - 9.8 m/s2 or – 32 ft/s2. (The negative sign means that gravity always acts downward)

• As we move above the Earth’s surface or to other planets, the strength of the gravitational field changes and so does the weight.

Planet

Force of gravity

Orb

ital

path

Force of gravity (weight) at the Earth’s surface

Fearth = G m1 me

re 2

Force of gravity (weight) at a point (P) above the

Earth’s surface

Fpoint P = G m1 me

rp2

w = m1g = G m1 me

r2

g = G me

r2

g r2 = Gme

gearth r2earth = Gmearth

gpoint p r2point P = Gmearth

Therefore

gearth r2earth =gpoint p r2

point P

Radius of Earth = 4000 miles

scale150 lbs

Two Radius of Earth = 8000 miles

scale37.5 lbs

Three Radius of Earth = 12000 miles

scale16.7 lbs

¼ wt

1/9 wt

Normalwt

scale

150 lbs

scale

25.6 lbs

scale

406 lbs

g = 9.81 m/s2 g = 1.67 m/s2 g = 26.6 m/s2

Satellites

• When satellites orbit a planet the force which supplies the centripetal force, and thereby the circular motion, is the pull of gravity of the planet which is orbited.

• Without the force of gravity, the satellite would move in a straight line due to inertia.

• When the satellite orbits, the force of gravity must equal the centripetal force. If the force of gravity exceeded the centripetal force the satellite would spiral into the planet. If the centripetal force exceeded the force of gravity, the satellite would seek a wider orbit or move off in a straight line.

VelocityVectors

AccelerationVectors

ForceVectors

Earth

A satellite is a projectile shot from a very high elevationand is in free fall about the Earth.

Inertial position

Centripetal force

Centripetal force

Centripetal force

Centripetal force

Gravity suppliescentripetal forceinward towardsthe center of the

circular path

Fg = gravity force betweenm1 and m2 separated by

a distance r

G is the Universal Gravitational Constant

The weight of an object is its mass times g’, the

gravity value at location r

Planet

Force of gravityFg

Centripetal forceFc

Orb

ital

path

Fg = Fc

Fg = G m1 m2

r2

Fc = m v2

r

G m1 m2 = m1 v2

r2 rCanceling m1 & r on both sides

V2 = G m2

r

(1) V2 = G m2

r (2) V2 r= G m2

(3) V = ωr (4) ω= 2πf (5) T = 1/f

(6) ω = 2π / T (7) (ωr)2r = Gm2

(8) ω2 r3 = Gm2

(9) ( 2π / T)2 r3 = Gm2

(10) 4π2 r3 / T2 = Gm2

(11) T2 / r3 = 4π2/ Gm2 = a constantT2 / r3 = a constant

Kepler’sThird Law

Kepler’s Laws of Satellite Motion• (1) Satellites travel in elliptical paths. (The Earth and

the inner planets as well as the moon travel in nearly circular orbits. The orbits of the outer planets are more ellipsoid. Comets orbits are very elliptical.)

• (2) Areas swept out in equal times are equal even though the speed of the satellite varies. Satellite velocity is least when it is furthest from the central body (apogee) and greatest when it is nearest (perigee).

• (3) The period of motion squared divided by the average orbital radius cubed gives a constant for all satellites orbiting the same body. ( T2

1/ r31 = T2

2/ r32)

Velocity increases(perigee)

Velocitydecreases(apogee)

Equal AreasIn

Equal Times

FORCE

(N)

DISPLACEMENT (M)

X1 X2

WORK = AREA UNDER THE CURVEW = F X (SUM OF THE BOXES)

WIDTH OF EACH BOX = X

AREA MISSED - INCREASINGTHE NUMBER BOXES WILL

REDUCE THIS ERROR!

AS THE NUMBER OF BOXESINCREASES, THE ERROR

DECREASES!

WEIGHT

(N)

Distance Above Center of Earth (m)

r1 r2

WORK = AREA UNDER THE CURVEW = F X (SUM OF THE BOXES)

r1 & r2 are are two pointsin the gravity field

Law of Universal Gravitation

Fg = G m1 m2

r2

Work = Fdr = G m1 m2 dr r2

Work = - G m1 m2

r

r2

r1

|

Work = - G m1 m2

r

r2 = infinityr1 = radius of the Earth

K.E. work of lift satelliteK.E. = ½ mv2

½ m1 v2 = - G m1 m2

r

Vesc = 2 G m2 1/2

r

r2

r1

|

Velocity required to leave theEarth’s gravity field

)(

Gravitation & Satellite Problems(a)What is the velocity of a satellite 5000 km above

the Earth? (b)What is it period of rotation ?

5000 km

Re

r = re + hme = 6 x 10 24 kgRe = 6.4 x 10 6 m

(a) V2 = G m2 , v = G me 1/2

r rV = (6.67 x 10-11 )(6 x 1024) 1/2

(6.4 x 106 + 5 x 106)V = 5900 m/s

(b) V = 2π r / T, T = 2π r/ v T = 2π (11.4 x 106) / 5900 = 1.2 x 104 sec = 3.4 hr.

( )( )

Gravitation & Satellite ProblemsWhat is the gravitational force between two elephants which

are 2 meters apart. Each elephant weighs 3200 lbs.

• m = 3200lbs / 2.2 lbs/kg = 1455 kg (now using MKS units and the Law of Universal Gravitation)

• Fg = G m1 m2 = 6.67 x 10-11 (1455)(1455)

r2 22

• Fg = 3.5 x 10-5 N

Fg = G m1 m2

r2

W = mg

Gravitation & Satellite ProblemsFind the escape velocity of an object leaving the Earth (escape velocity is the velocity required for an object to

leave Earth’s gravity as a projectile).

• mearth = 6 x 1024 kg, rearth = 6.4 x 106 m, G = 6.67 x 10-11 N m2 / kg2

• V = 2 (6.67 x 10-11 )(6 x 1024 ) 1/2

(6.4 x 106 )• V = 1.12 x 104 m/s or 25,000 mph

Vesc = 2 G m2 1/2

r

Escape velocity formula

( )

( )

Gravitation & Satellite ProblemsFind the height of a geosynchronous Earth satellite. (A satellite which maintains its position at a fixed point above the Earth)

• Using Kepler’s third law

• And the facts that the moon is 242,000 miles (3.9 x 108 m) from Earth and its period is 27.3 days (2.36 x 106 sec)

• (27.3)2 / (242,000)3 = (1)2 / rsat3,

• rsat = ((12 x 242,0003) / 27.32))1/3 = 26,700 miles above Earth’s center or (26,700 – radius of Earth (4000 miles)) = 22,700 miles above the Earth’s surface.

In order to maintain its position theperiod of the satellite must equal that

of the Earth (24 hours or 1 day)

T21/ r3

1 = T22/ r3

2

Gravitation & Satellite Problems(a) Find the acceleration due to gravity at 1000 km

above the Earth. (b) What is the weight of a 70 kg man at this location ?

• (a) gearth r2earth =gpoint p r2

point P

• (9.8) x (6.4 x 106 )2 = gp (6.4 x 106 + 1 x 106)2

• gp = 7.33 m/s2

• (b) g at 1000 km above the Earth is 7.33 m/s2 and weight = mass x gravity, therefore• w = 70 kg x 7.33 m/s2 = 513 N

1000 km

Re

gearth r2earth =gpoint p r2

point P

rearth = 6.4 x 106 mgearth = 9.8 m/s2

What is the force of attraction between two 1000 kg objectsSeparated by 3 meters? Express the answer in newtons.

(A) 0.13 (B) 0.0000074 (C) 0.000032 (D) 320

Find the value of gravity at 2000 km above the Earth? Express the answerin meters per second squared.

(A) 32 (B) 5.7 (C) 16.9 (D) 0

A person weighs 128 lbs on Earth. What is his weight in lbs, at 6.4 x 106

meters above the Earth ?(A) 64 (B) 10.2 (C) 32 (D) weightless

Find the radius of the orbit of a satellite above the Earth with a periodof 12 hours. Express the answer in kilometers

(A) 3200 (B) 2670 (C) 7200 (D) 320

A satellite orbits the Earth with a radius of 8000 km. Find its velocityIn meters per second.

(A) 7000 (B) 5000 (C) 550 (D) 20,000

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