1. Universal Law of Gravity

2. 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. 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.

3. Universal Law of Gravity Force of gravity has magnitude given by (Gravity Force) = (G) x (Gravity Force) = (G) x Object A Object B ( Mass of Object A ) x ( Mass of Object B) ( Distance ) x ( Distance ) DISTANCE Force Equal and opposite forces (Newton’s Third law)

4. Universal Gravity Constant, G In the formula for gravity force, we have G = 0.0000000000667 N m 2 / kg 2 = 6.67 x 10 –11 N m 2 / kg 2 The formula and the constant are called “universal” because, up to now, this theory predicts gravity anywhere in the universe.

5. Sample Problem Here is an example of using the formula (Gravity Force) = (G) x (Gravity Force) = (G) x ( Mass of Object A ) x ( Mass of Object B) ( Distance ) x ( Distance ) DISTANCE = Earth’s Radius Force Object B (Earth) Object A (1 kg mass)

6. Sample Problem Find gravity force for a 1 kg mass on surface of Earth. (Force) = (6.67 x 10 –11 ) x Value comes out to 9.8 Newtons ( 1 ) x ( 6 x 10 24 ) ( 6.38 x 10 6 ) 2 Universal Gravity Constant, G Earth’s Radius Earth’s Mass

7. Sample Problem (cont.) Find gravity acceleration on a 1 kg mass. Using Newton’s Second Law, (Acceleration) = = Answer is 9.8 m/s 2. We just confirmed our value of a!!! (Force) (Mass) ( 9.8 N ) (1 kg )

8. Sample Problem Are you attracted to the person sitting next to you? Calculate the gravitational attraction between you (assume 70 kg) and the person next to you (assume 65 kg) if you are 1.2 m apart. Are you attracted to the person sitting next to you? Calculate the gravitational attraction between you (assume 70 kg) and the person next to you (assume 65 kg) if you are 1.2 m apart. givensformula substitutionunknownFg

9. Answer = 2.11 x 10 -7 N Answer = 2.11 x 10 -7 N 0.000000211 N Very Small!!! But anything with mass has an attractive force. Very Small!!! But anything with mass has an attractive force.

10. Sample Problem Determine the force of gravitational attraction between the earth (m = 5.98 x 10 24 kg) and a 70- kg physics student if the student is standing at sea level, a distance of 6.38 x 10 6 m from earth's center. Determine the force of gravitational attraction between the earth (m = 5.98 x 10 24 kg) and a 70- kg physics student if the student is standing at sea level, a distance of 6.38 x 10 6 m from earth's center.

11. Newton and the Moon Newton realized that Earth’s gravity was the centripetal force that kept the moon in orbit. Also discovered that gravity was weaker at that great distance. Gravity force

12. Gravity & Distance We don’t notice that gravity gets weaker as we move away from Earth because we rarely go very far. Moon is 30 Earth diameters away

13. Value of g (acceleration due to gravity; m/s 2 ) Using F grav = m·g we can derive: Using F grav = m·g we can derive:

14. Value of g Can now find g anywhere Can now find g anywhere g = Gm/r 2 g = Gm/r 2 Where Where G = 6.67 x 10 -11 N m 2 /kg 2 m = mass of planet in kg r = radius of planet in meters

15. Known as “Inverse Square Law” Gravity force weakens with distance as the inverse of the square of the distance. Geometric property of area and distance. outer circle is twice Earth’s radius Earth Gravity 1/4 Earth Gravity

16. Sample Problem Find your weight if you have a mass of 60 kg and are 2.1 x 10 5 m above the earth’s surface. Find your weight if you have a mass of 60 kg and are 2.1 x 10 5 m above the earth’s surface.

17. Solution Givens: r = 2.1 x 10 5 m + 6.37 x 10 6 m = 6.58 x 10 6 m m you = 60 kg m earth = 5.98 x 10 24 kg Unknown: Equations: F g F = ma for weight F g = mg Fg = Fg = (60kg) (9.2 m/s 2 ) Fg = Fg = (60kg) (9.2 m/s 2 ) mg = G m 1 m 2 = 553 N r 2 g = G m g = G m r 2 r 2 g = 9.2 m/s 2 g = 9.2 m/s 2

18. Weightlessness In deep space, far away from all stars, planets, etc. there is almost no gravity force. In orbit near Earth, gravity is still strong (only 10% less than on surface). Why are Shuttle and Space Station astronauts “weightless”? Earth is nearby

19. NASA’s “Vomit Comet” NASA has a special airplane for training astronauts in free-fall weightless conditions. The “Vomit Comet” nickname tells you it’s quite a wild roller- coaster ride. The plane flies between 20,000 and 30,000 feet, same as commercial flights.

20. Power Climb Weightless Freefall Pull out of Dive Flight of the “Vomit Comet” At the top of the arc, the plane’s trajectory is projectile motion. Boeing 707 (modified)

21. Orbits Geosynchronous Orbit – orbits above the same point on the equator of the earth at all times Geosynchronous Orbit – orbits above the same point on the equator of the earth at all times GPS, cell phones, etc. GPS, cell phones, etc.

22. Orbits and Centripetal Force Gravity provides the centripetal force required for a satellite to move in a circle. F g = F c F g = F c mg = mv 2 /r g = v 2 /r g = v 2 /r

23. Sample Problem Calculate the speed needed for one of the Direct TV satellites to orbit at an altitude of 320,000 m. Calculate the speed needed for one of the Direct TV satellites to orbit at an altitude of 320,000 m.

24. Getting into Orbit Rocket needs to lift above the atmosphere and then fire thrusters to acquire the required orbital speed of about 8 kilometers per second. Returning to Earth, air resistance slows the spacecraft during reentry.

25. Elliptical Orbits For speeds higher than 8 km/s, the orbit is elliptical instead of circular.

26. Escape Speed If speed exceeds 11.2 km/s then object escapes Earth because gravity weakens (as object gets further away) and never slows the object enough to return it back towards Earth. Circular Elliptical Hyperbolic

27. Using the Law of Universal Gravitation Newton’s Law is used and applied to the motion of the planets about the sun. Newton’s Law is used and applied to the motion of the planets about the sun. Newton derived an equation from the Law of Universal Gravitation to describe the Period of Planetary Motion: Newton derived an equation from the Law of Universal Gravitation to describe the Period of Planetary Motion: T² = (4π²/Gm s )r³