The moon revolves around the earth making the complete revolution in 27.3 days assume that the orbit is circular and has a radius r=238000 miles. What is the magnitude of the gravitational force exerted on the moon by the earth
Answers
Answer:
hopw it helps
Explanation:
An object moving in a circle of radius r with constant speed v is accelerating. The direction of its velocity vector is changing all the time, but the magnitude of the velocity vector stays constant. The acceleration vector cannot have a component in the direction of the velocity vector, since such a component would cause a change in speed. The acceleration vector must therefore be perpendicular to the velocity vector at any point on the circle. This acceleration is called radial acceleration or centripetal acceleration, and it points towards the center of the circle. The magnitude of the centripetal acceleration vector is ac = v2/r. (We skip the derivation of this expression.)
Link: Uniform circular Motion
Problem:
The orbit of the moon around the earth is approximately circular, with a mean radius of 3.85*108 m. It takes 27.3 days for the moon to complete one revolution around the earth. Find
(a) the mean orbital speed of the moon and
(b) its centripetal acceleration:
The gravitational force is relatively simple. It is always attractive, and it depends only on the masses involved and the distance between them. Stated in modern language, Newton’s universal law of gravitation states that every particle in the universe attracts every other particle with a force along a line joining them. The force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
Gravitational attraction is along a line joining the centers of mass of these two bodies. The magnitude of the force is the same on each, consistent with Newton’s third law.
