1 / 0

Newton’s Law of Gravity

Newton’s Law of Gravity. Starter. To increase the speed of an Earth satellite, should it be placed in an orbit closer to the Earth or further away from the Earth?. Newton’s Law of Gravity. Starter. Newton’s Law of Gravity. Starter.

krikor
Download Presentation

Newton’s Law of Gravity

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Newton’s Law of Gravity Starter To increase the speed of an Earth satellite, should it be placed in an orbit closer to the Earth or further away from the Earth?
  2. Newton’s Law of Gravity Starter
  3. Newton’s Law of Gravity Starter If the mass of the Earth stayed the same but the radius of the Earth doubles, what happens to your weight? a. it doubles also. b. it stays the same. c. it drops to 1/2 of the original. d. it drops to ¼ of the original. g = MG/R2
  4. Practice Newton’s Law of Gravity : Every two objects attract each other with a gravitational force given by: F = m1m2G/r2 m1 = mass of the first object in kg m2 = mass of the second object in kg r = distance between the two masses in meters G = 6.67 x 10-11
  5. Practice / Example Find the gravitational force between the Earth and a 50,000kg asteroid that is 12 x 106 m away. m1 = 5.00 x104 m2= 6.00 x 1024 G = 6.67 x 10-11 r = 12 x 106 F = m1m2G/r2 =(5.00 x104)( 6.00 x 1024)(6.67 x 10-11 ) / (12 x 106)2 = 1.39 x 105 Newtons = 139,000 Newtons
  6. Where does g come from? Consider a person, mass m, standing on the Earth, mass M. F = mMG/R2 = mg g = MG/R2 = (6x1024)(6.67x10-11)/(6.4x106)2 = 9.8 m/s2
  7. Circular Orbits F = mv2/R = mMG/R2 and v = 2pR / T Orbit Velocity Period R = Orbit Radius M = mass of planet or sun being orbited (The mass of the satellite is not in these 2 equations.)
  8. Example The moon is 385 x 106 m from the Earth’s center. R = 385 x 106 MEarth = 6 x 1024 What is its orbit velocity? v = (MG/R)1/2 = ( ( 6 x 1024)(6.67 x 10-11)/(385 x106))1/2 = 1019 m/s What is the period? T = 2pR/v = (2p )(385 x 106)/ 1019 = 2.37 x 106 seconds 2.37 x 106seconds ( 1hr/3600s)(1 day/24 hr) = 27.5 days
  9. Useful Data
More Related