## Saturday, December 17, 2016

### Gravitation Problems with Solutions One

Gravitational force exists between every two particles having some mass and it is directly proportional to the product of their masses and inversely proportional to the square of distance of separation. It is independent of medium between them. If number of bodies are present around any body, the total gravitational force is the vector sum of all the existing forces. We also need to know that the presence of third body is not going to effect the gravitational force between the other two bodies. Acceleration due to gravity varies basing on the position of the body from the surface of the earth and depth of the body from the surface of the earth. It also depends on the location of the body with respect to its location of equator.

Problem

It is given in the problem that the gravitational force between two objects at a separation of one meter is given to us and we need to find the mass of each object.

Solution

We can apply law of gravitational force to know the force between the two bodies and the force is given in the problem and their separation is also given in the problem. By applying them in the formula and by applying the constant value of universal gravitational constant, we can solve the problem as shown in  the diagram below.

Problem

It is given in the problem that mass of one ball is four times the mass of the other body and their separation is given as 10 centimeter. The gravitational force between them is given to us and we need to find the mass of each particle.

Solution

We need to apply once again newton's law of gravitation as shown in the diagram below. Universal gravitational constant is a proportional constant and it is constant over the entire universe. Data can be further simplified and problem can be solved as shown in the diagram below.

Problem

It is given in the problem that three spherical balls of mass 1,2 and 3 kilogram are placed at the corners of an equilateral triangle of side one meter. We need to find the effective gravitational force acting on the body of mass one kilogram.

Solution

Let us consider three particles at the three corners of the equilateral triangle as shown in the diagram below. On one kilogram  particle there is now gravitational attractive force acting due to other two bodies of masses 2 and 3 kilogram. Their directions are along the line joining them and their resultant can be found using the parallelogram law of vectors.

It can be further simplified as shown in the diagram below. Vector law of addition is applied and we know the angular separation between the bodies and it can be further simplified as shown below.

Problem

We need to know that how a particle of known mass has to be divided into two parts such that the gravitational force between them is maximum and the problem is as shown in the diagram below.

Solution

We can find the gravitational force between any two bodies using the Newton's law of gravitation. We also need to know that if any function has to be maximum, its corresponding differentiation has to become zero as shown in the diagram below. By applying the rules of differentiation and simplifying it further, we can solve the problem as shown in the diagram below.

Problem

It is given that two particles of known mass are separated by a certain distance. We need to know at what distance from the body of mass one kilogram, the gravitational force becomes zero and the problem is as shown in the diagram below.

Solution

We shall imagine any particle of certain mass between the two bodies along the line joining them. On the third body, both the bodies applies force of attraction towards them and hence both of them on the third body are in opposite direction. At any instant, the resultant force becomes zero and the point becomes a neutral point. The problem is solved as shown in the diagram below.

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