1. Universal Gravitation

2. 2/28 Opener A 1. 2 kg bucket is whirled in a vertical circle with r=1
2/28 Opener A 1.2 kg bucket is whirled in a vertical circle with r=1.3m the speed of the bucket is 3.7 m/s at the top of the circle. Draw a freebody diagram. And determine Fgrav, Ftens, a, and Fnet.

3. What do we know about gravity so far?
Turn to your neighbor and share your knowledge of gravity from this class and other experiences.

4. Inverse Square Law
The force of gravity between the earth and any object is inversely proportional to the square of the distance that separates the object’s center from the earth’s center.
Fgrav~1/d2
Fgrav represents the force of gravity between the two objects.
How are force and distance related?
What happens if distance is doubled?
What happens if distance is halved?

5. Practice Question
Suppose that two objects attract each other with a gravitational force of 16 units. If the distance between the two objects is doubled, what is the new force of attraction between the two object? 4

6. Plicker Question
Suppose that two objects attract each other with a gravitational force of 16 units. If the distance between the two objects is tripled, what is the new force of attraction between the two object?
A Units
B Units
C Units
D. 4 Units
1.77 (/9)

7. Plicker Question
Suppose that two objects attract each other with a gravitational force of 16 units. If the distance between the two objects is halved, what is the new force of attraction between the two object?
A. 8 Units
B. 64 Units
C Units
D. 4 Units 64

8. Plicker Question
Suppose that two objects attract each other with a gravitational force of 16 units. If the distance between the two objects is reduced by a factor of 5, what is the new force of attraction between the two object?
A. 3.2 Units
B Units
C. 25 Units
D Units
400

9. Newton’s Law of Universal Gravitation
All objects attract each other with a force of gravitational attraction. Gravity is universal.
This force of gravitational attraction is directly dependent upon the masses of both objects and is inversely proportional to the square of the distance that separates their centers.
Fgrav = m1m2/d2
Fgrav represents the force of gravity between objects
mx represents the mass of each object
d represents the distance separating the objects
Talk with you neighbor how does mass impact gravitational force between objects?

10. 3/1 Opener
Suppose that two objects attract each other with a gravitational force of 100 units. If the distance between the two objects is 5 times greater, what is the new force of attraction between the two object? If the distance between the two objects is halved, what is the new force of attraction between the two object?

11. Universal Gravitation Constant (G)
G= x Nm2/kg2
When G is multiplied by Newton’s Universal Gravitation Equation units = N
Therefore Fgrav = G×m1×m2/d2
Plicker Question: The more massive that an object is, the ______ that the object will be attracted to Earth.
More B. Less C. Nonsense!
Plicker Question: The more the massive Earth is, the ______ that the object will be attracted to Earth.
Plicker Question: The the greater the earth’s radius, the ______ that the object will be attracted to Earth.
more

12. Practice Problem
Determine the force of gravitational attraction between the earth (m=5.98 x 1024 kg) and a 70 kg physics student standing at sea level 6.38 x 106m from the earth’s center.
686N

13. Plicker Problem
Determine the force of gravitational attraction between the earth (m=5.98 x 1024 kg) and a 70 kg physics student in a plane 4000 m from the earth’s surface (which is 6.38 x 106 m from the earth’s center).
A. 684N
B. 6.6 x 1012
C. 2.6 x 1016
D. 9.3 x 1017
684N

14. Short Cut
Determine the force of gravitational attraction between the earth (m=5.98 x 1024 kg) and a 70 kg physics student in a plane m from the earth’s surface (which is 6.38 x 106 m from the earth’s center).
Based on our answers a change of only 10000m from the center of the earth is negligible.
Therefore Fgrav = m x g = 70 x 9.8 = 686N

15. Practice Calculations

16. Cavendish Experiment
Experimentally determined the Universal Gravitation Constant (G= 6.673x10-11)
The small value reflects that it is only substantial for objects with large mass.
Cavendish was able to use the torsion balance to determine G with measured values of m1, m2, d, and Fgrav

17. Calculating g g represents the acceleration of gravity
g varies inversely with the distance from the center of the earth. It follows the inverse square law.
To find the acceleration of gravity on the surface of other planets g= G x mplanet/Rplanet2
We can prove g on earth’s surface is 9.8
9.8=(6.673 x 10-11) x (5.98 x 1024) / (6.38 x 106)2

18. Practice Problems YAY!
9.8,9.5,.22,