Presentation is loading. Please wait.

Presentation is loading. Please wait.

Physics 111 SI.  F= -G ( m1 * m2 / r^2 )  G = 6.67*10^-11 N.m^2/kg^2  g = G ( Me / Re^2) = G ( Me / ( Re +h)^2)  W = G ( Me * m / Re^2)  V = (G *

Similar presentations


Presentation on theme: "Physics 111 SI.  F= -G ( m1 * m2 / r^2 )  G = 6.67*10^-11 N.m^2/kg^2  g = G ( Me / Re^2) = G ( Me / ( Re +h)^2)  W = G ( Me * m / Re^2)  V = (G *"— Presentation transcript:

1 Physics 111 SI

2  F= -G ( m1 * m2 / r^2 )  G = 6.67*10^-11 N.m^2/kg^2  g = G ( Me / Re^2) = G ( Me / ( Re +h)^2)  W = G ( Me * m / Re^2)  V = (G * Me / r ) ^1/2  T= 2 пr ^ 3/2 / ( G * Me)^1/2  T is proportional to r^3/2  U = -G ( m * Me / r )  E= K + U =.5mv^2 + -G ( m * Me / r )  Conservation of Angular Momentum:  Ii Wi = If Wf  For elliptical orbits: ra va = rp vp

3 6. Planets A, B and C are identical. A and C have a giant moon orbiting them, while B has a lightweight artificial satellite orbiting it, as shown in the diagram. Which planet has the strongest gravitational interaction with its satellite? a. Planet A, because its moon is heavy and close to it. b. Planet B, because only a lightweight object can orbit without falling down. c. Planet C, because it can interact with a heavy object that is far away. d. All have the same gravitational attraction, because the planets are all the same mass. e. Not enough information to determine.

4  Why?  Use equation  F= -G ( m1 * m2 / r^2 )  Plant A has the shortest r and greatest m2 weight, producing the greatest Force value

5 Why doesn’t the gravitational pull between the Sun and the planets cause the planets to fall into the Sun? a. The background stars pull back on the planets. b. The Sun’s magnetic fields are pushing out. c. The speed of the planets flings them out. d. The Sun’s rotation pushes the planets out.

6  Satellites B is three times more massive than A, but orbiting the planet at three times the distance. a. A experiences a stronger force than B, because differences in distance are more influential than differences in mass. b. B experiences a stronger force than A, because differences in mass are more influential the differences in distance. c. A and B experience the same force, because differences in distance are exactly compensated by differences in mass. d. A and B experience no force, because they are in space.

7  a. A experiences a stronger force than B, because differences in distance are more influential than differences in mass.  Distance is squared, so it affects the force at an exponential rate, as an increase in mass progresses linearly

8  Planet A experiences a gravitational force by Planet C that is ­­­­­­­­­­­ ______ that by Planet B. a. less than b. equal to c. greater than d. cannot be determined with the given information.

9  Equal to!  A to B force:  F= -G ( m1 * m2 / r^2 )  F= -G ( 1 *1 / 1^ 2) = 1 A to C force: F= -G ( m1 * m2 / r^2 ) F= -F ( 1 * 4 / 2^2) = 4/4 = 1

10  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 in an airplane at 40000 feet above earth's surface. This would place the student a distance of 6.39 x 10 6 m from earth's center.

11  F= -G ( m1 * m2 / r^2 )  G = 6.67*10^-11 N.m^2/kg^2  F = (6.67*10^-11)(5.98 x 10 24 )(70) / (6.39 * 10 ^6)  F= 684 Newtons

12  Suppose that you have a mass of 70 kg (equivalent to a 154-pound person). How much mass must another object have in order for your body and the other object to attract each other with a force of 1- Newton when separated by 10 meters?

13  m = 2.14 x 10 10 kg  Use the equation F grav = G m 1 m 2 / d 2  where m 1 = 70 kg, d = 10 m and G = 6.673 x 10 - 11 Nm 2 /kg 2.  Substitute and solve for m 2. Note that the object is equivalent to an approximately 23 million ton object!! It takes a large mass to have a significant gravitational force.

14  The average orbital distance of Mars is 1.52 times the average orbital distance of the Earth. Knowing that the Earth orbits the sun in approximately 365 days, use Kepler's law of harmonies to predict the time for Mars to orbit the sun.

15  Given: R mars = 1.52 R earth and T earth = 365 days  Use Kepler's third law to relate the ratio of the period squared to the ratio of radius cubed  T is proportional to r^3/2  (T mars ) 2 / (T earth ) 2 (R mars ) 3 / (R earth ) 3 (T mars ) 2 = (T earth ) 2 (R mars ) 3 / (R earth ) 3  (T mars ) 2 = (365 days) 2 * (1.52) 3  (Note the R mars / R earth ratio is 1.52)  T mars = 684 days

16


Download ppt "Physics 111 SI.  F= -G ( m1 * m2 / r^2 )  G = 6.67*10^-11 N.m^2/kg^2  g = G ( Me / Re^2) = G ( Me / ( Re +h)^2)  W = G ( Me * m / Re^2)  V = (G *"

Similar presentations


Ads by Google