Universal Gravitation 228 Opener A 1 2 kg

Universal Gravitation

2/28 Opener A 1. 2 kg bucket is whirled in a vertical circle with r=1. 3 m 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.

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

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/d 2 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?

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?

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. 5. 33 Units B. 1. 77 Units C. 4096 Units D. 4 Units

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. 4096 Units D. 4 Units

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. 1048576 Units C. 25 Units D. 400 Units

Newton’s Law of Universal Gravitation o All objects attract each other with a force of gravitational attraction. Gravity is universal. o 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. o Fgrav = m 1 m 2/d 2 o o o Fgrav represents the force of gravity between objects mx represents the mass of each object d represents the distance separating the objects o Talk with you neighbor how does mass impact gravitational force between objects?

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?

Universal Gravitation Constant (G) G= 6. 673 x 10 Nm /kg -11 2 2 When G is multiplied by Newton’s Universal Gravitation Equation units = N Therefore Fgrav = G×m 1×m 2/d 2 Plicker Question: The more massive that an object is, the ______ that the object will be attracted to Earth. A. More B. Less C. Nonsense! Plicker Question: The more the massive Earth is, the ______ that the object will be attracted to Earth. A. More B. Less C. Nonsense! Plicker Question: The the greater the earth’s radius, the ______ that the object will be attracted to Earth. A. More B. Less C. Nonsense!

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 106 m from the earth’s center.

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. 684 N B. 6. 6 x 1012 C. 2. 6 x 1016 D. 9. 3 x 1017

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 10000 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 10000 m from the center of the earth is negligible. Therefore Fgrav = m x g = 70 x 9. 8 = 686 N

Practice Calculations

Cavendish Experimentally determined the Universal Gravitation Constant (G= 6. 673 x 10 -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 m 1, m 2, d, and Fgrav

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/Rplanet 2 We can prove g on earth’s surface is 9. 8=(6. 673 x 10 -11) x (5. 98 x 1024) / (6. 38 x 106)2

Practice Problems YAY!