Download presentation

Presentation is loading. Please wait.

Published byGervais Brown Modified about 1 year ago

1
“Concern for man himself and his fate must always form the chief interest of all technical endeavors… Never forget this in the midst of your diagrams and equations.” Albert Einstein

2
Gravity Math

3
The History We have learned the history of gravity, the missteps and false starts, the names of the movers and shakers in the field, and the progression of the ideas

4
The Math The math for gravity is a mix of the equations for circular motion and the physics for acceleration

5
What could we want to know? Let us make a list of the possible bits of information we could want to know about an object orbiting around another, or about the attraction between two objects

6
The List Constants Acceleration Force Period Velocity Mass of either object

7
The Constant There is one constant involved, the gravitational constant

8
Acceleration We know that a change in direction requires an acceleration We saw this for circular motion

9
Acceleration r is the distance from the center of the mass M is the mass of the attracting object

10
Note… The mass of the falling object is not a factor Only the mass of the planet, Sun, or other large, attracting body and distance of the item from the center counts

11
Example 1 At sea-level the radius of the Earth is 6.38 × 10 6 m. If the mass of the Earth is 5.98 × 10 24 kg, what is gravitational acceleration at sea-level? What is it atop Mt. Everest, 8848 m above sea-level?

12
What Do We Know G = 6.67 × 10 -11 Nm 2 /kg 2 r 1 = 6.38 × 10 6 m m = 5.98 × 10 24 kg r 2 = ((6.38 × 10 6 )+ 8848) m

13
Answer 1 a = Gm/r 2 a 1 = 9.799 m/s 2 ≈ 9.80 m/s 2 a 2 = 9.772 m/s 2 ≈ 9.77 m/s 2

14
Example 2 Jupiter orbits the sun at a radius of 7.78 ×10 11 m. The Sun has a mass of 1.99 ×10 30 kg. What is the acceleration holding Jupiter in its orbit?

15
What Do We Know G = 6.67 × 10 -11 Nm 2 /kg 2 r = 7.78 ×10 11 m m = 1.99 ×10 30 kg

16
Answer 2 a = Gm/r 2 a = 2.1929 × 10 -4 m/s 2 ≈ 2.19 × 10 -4 m/s 2

17
Force We already know the basic equation

18
Force If we know the mass of the object, we need only find the acceleration to calculate the force

19
Force

20
G is the gravitational constant M is the larger of the two masses m is the smaller of the two r is the radius between them

21
Force This was Newton’s big achievement He was able to find out what Kepler’s constant was

22
Example 3 Pluto has a mass of 1.25 × 10 22 kg and an orbital radius of 5.87 × 10 12 m. What is the force, in Newtons, between Pluto and the Sun?

23
What Do We Know G = 6.67 × 10 -11 Nm 2 /kg 2 r = 5.87 × 10 12 m m = 1.25 ×10 22 kg M = 1.99 ×10 30 kg

24
Answer 3 F = GMm/r 2 F = 4.82 × 10 16 N

25
Period

26
Remember, this is the time it takes to make one full revolution The time is measured in seconds

27
Example 4 What is the orbital period of Pluto, given the information from the pervious problems?

28
What Do We Know G = 6.67 × 10 -11 Nm 2 /kg 2 r = 5.87 × 10 12 m M = 1.99 ×10 30 kg

29
Answer 4 T = 2π(r 3 /(GM)) 1/2 T = 7.76 × 10 9 s Converted to years T = 246 years

30
Example 5 The Earth has a period of 365 days. If it orbits at a radius of 1.50×10 11 m, then what is the mass of the Sun.

31
What Do We Know G = 6.67 × 10 -11 Nm 2 /kg 2 r = 1.50 × 10 11 m T = 365 days = 3.15 × 10 7 s

32
Answer 5 T = 2π(r 3 /(GM)) 1/2 M = (4π 2 r 3 )/(GT 2 ) M = 2.01 × 10 30 kg

33
Velocity

34
The velocity of the orbiting object depends only on the mass of the larger body, not the mass of the orbiting object

35
Example 6 A satellite orbits the Earth at a distance of 340000 m. The mass of the Earth is 5.97 × 10 24 kg. What velocity must it have to maintain that orbit?

36
What Do We Know G = 6.67 × 10 -11 Nm 2 /kg 2 M = 5.97 × 10 24 kg r = 340000 m

37
Answer 6 v = ((GM)/r) 1/2 v = 34222 m/s ≈ 34200 m/s

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google