Newton believed that every object ___attracts_____ every other object. The force of the attraction depends on the __mass___ and _distance__ of the two objects.

F g = G (m 1 m 2 ) r 2 F g = gravitational force in (N) m 1 = mass object 1 (kg) m 2 = mass object 2 (kg) r = the distance between the masses (m) (measured from their centers) G = 6.67 x N m 2 /kg 2 (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: m 1 = 70 kg m 2 = 65 kg r = 1.2 m G = 6.67 x N m 2 /kg 2 Unknown: F g Equation: F g = G (m 1 m 2 ) r 2 = (6.67 x N m 2 /kg 2 )(70 kg)(65 kg) (1.2 m) 2 = 2.11 x N = N (tiny!)

 As you go further away from the earth’s surface the acceleration due to gravity _decreases____.  Two equations for gravitational force F = ma becomes F g = mg and F g = G (m 1 m 2 ) r 2 We can set them equal to each other: F g = F g mg = G m 1 m 2 one of the masses cancels r 2 g = G m r 2

 g = the acceleration due to gravity (m/s 2 )  G = 6.67 x N m 2 /kg 2  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 10 5 m above the earth’s surface. Given: r = 2.1 x 10 5 m x 10 6 m = 6.58 x 10 6 m m earth = 5.98 x kg Unknown: F g Equation: F = ma, now F g = mg ( We need g at that altitude.) F g = F g mg = G m 1 m 2 r 2 g = G m r 2 = (6.67 x N m 2 /kg 2 )(5.98 x kg) (6.58 x 10 6 m) 2 = 9.21 m/s 2

So you would weigh F g = mg = (60kg) (9.21 m/s 2 ) F g = 553 N

A geosynchronous satellite orbits above the same point on the equator of the earth at all times. Examples: GPS cell phone satellites TV satellites

Satellites are _____projectiles_____. 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 gravitational force = centripetal force F g = F c mg = m v 2 r (The mass of the satellite cancels out!) g = v 2 r g = the acceleration due to gravity in m/s 2 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 x 10 6 m = 6,690,000 m or 6.69 x 10 6 m Unknown: v Equation: F g = F c mg = mv 2 r g = v 2 r We need g at that altitude: F g = F g mg = G m 1 m 2 r 2 g = G m r 2 = (6.67 x N m 2 /kg 2 )(5.98 x kg) (6.69 x 10 6 m) 2 g = 8.91 m/s m/s 2 = ___v 2 __ (6.69 x 10 6 m) v = 7,721 m/s (that’s about 17,000 mph!)