Gravitational field intensity is the force experienced by a unit massed particle when it is placed in the gravitational field of another particle of known mass and its direction is similar to gravitational force. Planetary motion is identified basing on Kepler laws. According to first law, planets take elliptical path around a star. According to second law, areal vector of a planet sweeps equal area in in equal intervals of time. Basing on this it can be said that when a planet comes near to the star its velocity will be more and vice versa.

**Problem**

Infinite number of masses each of mass one kilogram are placed along a straight line at a uniform distance of separation from the origin. We need to find the gravitational intensity at the origin due to all the other masses and the problem is as shown in the diagram below.

**Solution**

We know that gravitational force acts on the particle at the origin and as we need to find the intensity, we shall consider a unit mass at the origin. Each particle on the x axis applies gravitational force of attraction on it and the total force is the vector sum of all the forces acting on it. Solution is as shown in the diagram below. Sum of the terms can be found using the geometrical progression sum using the standard formula as shown in the diagram below.

**Problem**

Gravitational field due to a mass at a distance due to a given mass as shown in the equation below. Assuming gravitational potential to be zero at infinity, we need to find its value at a given point and the problem is as shown in the diagram below.

**Solution**

We can express the gravitational potential in terms of the intensity as shown in the diagram below. Potential is the work need to be done to bring a mass of unit mass from infinite to a particular location. Problem can be further simplified as shown in the diagram below.

**Problem**

Two spheres are separated by a known distance and we need to know the angle which the strings will make with the vertical due mutual attraction of the spheres and the problem as shown in the diagram below.

**Solution**

As the strings turn tight, tension is developed in the string and it can be resolved into components. By dividing the two equations, we can solve the problem as shown in the diagram below.

**Problem**

The value of acceleration due to gravity on the surface of the earth is given. If the point where we need to find the acceleration due to gravity from the surface of the earth and we need to find that height and the problem is as shown in the diagram below.

**Solution**

We can just apply the acceleration due to gravity variation with respect to height and simplify the equation as shown in the diagram below.

**Problem**

Three particles are placed at the three corners of a equilateral triangle of side a. We need to find the work done in increasing the side length to double to its initial value. Problem is as shown in the diagram below.

**Solution**

Each particle will get some potential energy with respect to other and as the separation increases potential energy also increases. Work done is the difference between initial and final potential energies of the system. Solution is as shown in the diagram below.

**Related Posts**

**Resultant Gravitational Force and Neutral Point**

**Gravitational Potential Energy, Kinetic energy and Total Energy**

**Orbital Velocity and Escape Velocity**

Good collection and better explanation.

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