## Sunday, December 18, 2016

### Gravitation Problems with Solutions Two

We are  solving series of problems about a physics concept called gravitation. We are predominately are solving problems based on force of attraction between two bodies. It is always attractive nature and it is never repulsive. This force exists between every two bodies with out any escape as they are all having some masses. It is independent of medium and the proportionality constant is called universal gravitational constant. The variation of acceleration due to gravity varies with height and depth from the surface of the earth. The loss of acceleration due to gravity is quicker with height when compared with the depth.

Problem

Here in this problem we need to know the variation of acceleration due to gravity with the depth from the surface of the earth. Problem is as shown in the diagram below.

Solution

Let us consider the location where we need to measure the acceleration due to gravity is at a depth d from the surface of the earth. From the center of the earth then it is at a distance of R-d. So on the given mass, entire mass of the earth is not applying the gravitational force of attraction, rather a portion of the mass under it.

So we shall count only that portion of the mass. So we express it in terms of volume and density and further simplify as shown in the diagram below.

Problem

Three uniform spheres each of identical mass and radius are kept in such a way that each one touch the other two spheres. We need to find the total gravitational force acting on each sphere and the problem is as shown in the diagram below.

Solution

On any one sphere, there is gravitational force due to the other two and to find the resultant force, we need to use vector laws of addition. Distance between any two spheres centers is nothing but twice the radius of any sphere.

We can find the resultant as shown in the diagram below and its direction also can be identified. Complete and detailed solution is given below.

Problem

Four identical masses are placed at the four corners of a square and we need to find the effective gravitational force at any one corner and the problem is as shown in the diagram below.

Solution

On any one particle, there is gravitational force of attraction due to the other two. We can find each force using the newton's law of gravitation. The diagram is as shown below.

It can be further solved as shown below. Two of the total three forces are equal in magnitude and perpendicular to each other. Their resultant is along the direction of the third force and hence we can find the total resultant as shown in the diagram below.

Problem

A satellite is going along an elliptical path around the earth. We need to find the relation between rate of area swept by it in terms of its radius and the problem is as shown in the diagram below.

Solution

According to second law of Kepler, the velocity vector of a planet sweeps equal area in equal intervals of time. Taking that into consideration, we can solve the problem as shown in the diagram below.

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