Newton believed that every object ______________ every other object.

The force of the attraction depends on the __________ and ___________ of the two objects.

Fg = G (m1m2) r2 Fg = gravitational force in (N) m1 = mass object 1 (kg) m2 = mass object 2 (kg) r = the distance between the masses

(m) (measured from their centers)

G = 6.67 x 10-11 N m2/kg2

(a constant everywhere!)

Ex. 1 Are you attracted to the person sitting next to you? Calculate the gravitational force between you (mass 70kg) and the person next to you (mass 65kg) if you are 1.2 m apart.

Given: m1 = 70 kg m2 = 65 kg r = 1.2 m G = 6.67 x 10-11 N m2/kg2

Unknown: Fg Equation: Fg = G (m1m2) r2

= (6.67 x 10-11 N m2/kg2)(70 kg)(65 kg) (1.2 m)2

= 2.11 x 10-7 N

= 0.000000211 N (tiny!)

As you go further away from the earth’s surface the acceleration due to gravity _____________.

Two equations for gravitational force F = ma becomes Fg = mg and Fg = G (m1m2)

r2

We can set them equal to each other:

Fg = Fg

mg = G m1 m2 one of the masses cancels

r2

g = G m r2

g = the acceleration due to gravity

(m/s2) G = 6.67 x 10-11 N m2/kg2

m = the mass of the planet in kg r = the distance from the center of

the planet in (m)

Remember- r measures from CENTER of planet, not the surface.

Ex. 2 Find the gravity if you are 2.1 x 105 m above the earth’s surface.

Given: r = 2.1 x 105 m + 6.37 x 106 m = 6.58 x 106 m

mearth = 5.98 x 1024 kg Unknown: Fg

Equation: F = ma, now Fg = mg (We need g at that altitude.)

Fg = Fg

mg = G m1 m2

r2

g = G m r2

= (6.67 x 10-11 N m2/kg2)(5.98 x 1024 kg) (6.58 x 106 m)2

= 9.21 m/s2

So you would weigh Fg = mg

= (60kg) (9.21 m/s2)

Fg = 553 N

A geosynchronous satellite orbits above the same point on the equator of the earth at all times.

Examples:GPScell phone satellitesTV satellites

Satellites are _________________________. In order to not fall back to earth, they need to maintain a certain velocity…

In order for a satellite to stay in a consistent orbit __________________ = ________________

Fg= Fc mg = m v2 r (The mass of the satellite cancels out!) g = v 2 r

g = the acceleration due to gravity in m/s2

v = the speed of the satellite in m/s r = the distance from the center of the

planet in meters

Ex. 3 Calculate the speed needed for one of the Direct TV satellites to

orbit at an altitude of 320,000 m above the earth’s surface.

Given: r = 320,000 m + 6.37 x 106 m = 6,690,000 m or 6.69 x 106 m

Unknown: v Equation: Fg = Fc

mg = mv2

r g = v2 rWe need g at that altitude: Fg = Fg

mg = G m1 m2

r2

g = G m r2

= (6.67 x 10-11 N m2/kg2)(5.98 x 1024 kg) (6.69 x 106 m)2

g = 8.91 m/s2

8.91 m/s2 = ___v2__ (6.69 x 106 m)

v = 7,721 m/s (that’s about 17,000 mph!)