Reference no: EM131766331

According to that law, the magnitude of the gravitational force FG12 between two objects of masses m1 and m2, separated by a distance r, can be expressed with the following proportionality FG12∝m1m2r2

This applies not only to small particles, but also to large objects (e.g. planets and stars). In fact, the gravitational force between two uniform spheres is the same as if we concentrated all the mass of each sphere at its center. Thus, by modeling the Earth and the Moon as uniform spheres, you can use the particle approximation when calculating the force of gravity between them. Be careful in using Newton's law to choose the correct value for r. To calculate the force of gravitational attraction between two uniform spheres, the distance r in the equation for Newton's law of gravitation is the distance between the centers of the spheres. For instance, if a small object such as an elephant is located on the surface of the Earth, the radius of the Earth rEarth would be used in the law. Also note that in situations involving satellites, you are often given the altitude of the satellite, that is, the distance from the satellite to the surface of the planet; this is not the distance to be used in the law. There is a potentially confusing issue involving inertia and mass. While inertia refers to the tendency of an object to resist changes in motion, mass is a property that determines the gravitational force (and is proportional to the amount of matter contained by the object). As it turns out though, for normal, everyday speeds (i.e. nowhere near the speed of light in a vacuum!), we can treat these quantities as equivalent. Therefore, we will refer to the quantity denoted by m as mass in this tutorial as opposed to of inertia.

Part A

Two particles are separated by a certain distance. The gravitational force between them is F0. Now the separation between the particles is tripled. Find the new gravitational force F1. (Note: for the remainder of this tutorial, the "G" superscript in the term referring to the gravitational force will be omitted)

*answer in terms of F0.

F1 =

Part B

A satellite revolves around a planet at an altitude equal to the radius of the planet. The force of gravitational interaction between the satellite and the planet is F0. Then the satellite moves to a different orbit, so that its altitude is tripled. Find the new force of gravitational interaction F2.* answer in terms of F0.

F2 =

Part C

A satellite revolves around a planet at an altitude equal to the radius of the planet. The force of gravitational interaction between the satellite and the planet is F0. Then the satellite is brought back to the surface of the planet. Find the new force of gravitational interaction F4.

* answer in terms of F0.

F4 =

Part D

Two satellites revolve around Earth. Satellite A has mass m and has an orbit of radius r. Satellite B has mass 6m and an orbit of unknown radius rb. The forces of gravitational attraction between each satellite and Earth is the same. Find rb.

* answer in terms of r.

rb =

Part E

An adult elephant has a mass of about 5.0 tons. An adult elephant shrew has a mass of about 50 grams. How far r from the center of Earth should an elephant be placed so that the force of gravity exerted on it by Earth equals that of the elephant shrew on the surface of Earth? The radius of the Earth is 6400 km. (1ton=103kg.)

* answer in kilometers.

r =

 km