Showing posts with label Gravitation. Show all posts
Showing posts with label Gravitation. Show all posts

Gravitation Problems with Solutions Four

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.


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Gravitation Problems with Solutions Three

Gravitational force is one of the fundamental and weakest force of the nature and it exists between every two bodies as every body is having some masses. Gravitational force is due to the mass of the bodies and it is independent of the medium. It is a long distance ranged force and it acts between the particles separated by a very large distance. When compared with electromagnetic and nuclear force, it is just minimum and it gives the stability to the system. The  earth is revolving around the sun due to this gravitational force and it acts like the centripetal force. We are solving series of problems on this topic and this post is the third one in the series.

Problem

We would like to find the acceleration due to gravity of the earth at a height equal to half the radius of the earth above the earth. Problem is as shown in the diagram below.


Solution

While solving this case, approximation can not be applied as the height is significant when compared with the radius of the earth. By applying the newton's law of gravitation, we can find the gravity at a given height as shown in the diagram below.


Problem

Two equal masses are hung from a balance whose scale pans differ in height by h. If density of the earth is given to us, we need to find the error in the weighting. Problem is as shown in the diagram below.


Solution

It is shown in the diagram that from the reference point, both the pans are at different heights and hence they have a difference in the heights as given in the problem. Error in the reading is the difference in its weights. Variation of heights is small and hence approximation can be applied and the problem can be solved as shown in the diagram below.


Problem

A small body of super dense material whose mass is twice the mass of the earth but size is very small when compared with the earth. If it starts from the state of rest, we need to know its location after a specified time. The problem is as shown in the diagram below.


Solution

For every action there is equal and opposite reaction. As the body comes downward, the earth moves upward. The two meet at the center of mass of the system. We can further solve the problem as shown in the diagram below.


Problem

We need to find the height from the surface of the earth where weight of the body is similar to the weight at a depth from the surface of the  earth. Problem is as shown in the diagram below.


Solution

We need to apply the formula for the acceleration due to gravity at a height and at a depth and we need to just equate both of them. The problem is as shown in the diagram below.


Problem

We need to find how fast the earth has to rotate so that apparent weight of a body becomes zero at the equator. Problem is as shown in the diagram below.


Solution

We know how acceleration due to gravity varies with latitude. We know that at equator it is minimum and as we are moving to wards poles, it becomes maximum. By writing the formula at the equator and further equating weight at that point zero, we can find the angular velocity as seventeen times the normal velocity and the detailed solution is as shown in the diagram below.


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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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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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Orbital Velocity and Escape Velocity of a body from a planet

Orbital velocity

 Orbital velocity is the minimum velocity that has to be given to a body on a planet so that it can go to a certain height from the planet and there it starts rotating around that planet. Orbital velocity of a body around the planet depends on the mass of the planet and its radius. But it is independent of the mass of that body. It means the orbital velocity is same weather you want to orbit a small ball or a big and heavy body.

To determine orbital velocity of a body, we need to use the concept of centripetal force. Centripetal force is the resultant force acting on a body towards its center. It does not come from out side and the total effective force acting towards the center in a system is actually treated as centripetal force. When a body is revolving around the earth, centripetal force is nothing but the gravitational force between the body and the planet. By equating them, we can derive the equation for the orbital velocity of a body as shown in the video below.

For a body to revolve around the earth near to the earth, we shall give an orbital velocity approximately equal to 7.9 Km/sec. This is generally given to the artificial satellites send by us into the space. They further satisfy conditions like having a time period of 24 hours, rotates from west to east and if it is at a height of 36000 KM from the surface of the earth, they are called geostationary satellites.

 Orbital velocity is independent of the angle with which we project the body into the space.

Escape Velocity

Escape velocity is the minimum velocity that we shall give to a body on the planet so that in escapes away from the gravitational pull of the planet and never comes back to the planet. It goes into the space and gets away from the influence of the planet. This is also independent of the mass of the body that is escaping from the planet.

For the planet earth, escape velocity is approximately 11.2 Km/sec. It means if we give that much velocity to any body from the surface of the earth, it escapes away from the gravitational pull of the earth and never returns back to the earth.

It is proved in the following video lesson that the escape velocity is root two times the orbital velocity. It means 1.414 times more. It means for a body all ready orbiting around the earth, if you increase its speed by 0.414 or 41.4 %, it escape from the earth influence and goes to space.

What kind of gases present around a planet depends on its escape velocity and that escape velocity further depends on the mass of the planet and its radius.

For example RMS velocity of oxygen gas molecules is more than the escape velocity of the moon and hence all oxygen escaped from the moon. It could not do the same from the earth as the escape velocity of the earth is more than the RMS velocity of the oxygen gas. The same is the explanation for the hydrogen gas availability around the sun. The escape velocity of the sun is so large such that hydrogen cannot escape from its influence and acting like a basic source for nuclear fission in it.



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Gravitational Potential Energy, Kinetic Energy and Total Energy

Gravitational Potential energy

Gravitational Potential energy of a body on the earth Potential energy is the possessed by the body due to its position on the earth. The potential energy of a body can be measured taking the center of the earth as the reference. It means if the body is at the center of the earth, its potential energy is treated as zero.

Potential energy of the body is also treated as negative and it is a symbolic way of identifying that the body is associated with the system. We need to do some work it getting a body to a particular position on the earth. After the work is done, the energy used in doing the work cannot disappear and it will convert into some other form of energy. This is conservation of energy and energy is neither created and nor created. It just converts from one form to other. Thus the work done by us in putting the body at a certain place is converted into potential energy and it is called gravitational potential energy and it is by default treated as negative.

We can measure the work done in pulling the body to a certain mass by a small distance using the formula that the work done is the product of force and displacement. Force between the body and the mass of the earth can be measured using Newton’s law of gravitation.

In this way, we are going to get the work done in displacing the body only by a small value and we need to know the work done in producing the total displacement. This can be done by integrating the first equation. This total work done is stored in the form of gravitational potential energy.

As we move far away from the surface of the earth, potential energy starts increasing. At infinite height, potential energy becomes zero. Thus potential energy of a body associated with the earth is negative and at the best, at infinity, it will become zero and it cannot turn into a positive value.



Change in Potential Energy

When the body is on the surface of the earth, it will possess some potential energy. When it is moved to a certain height, its potential energy increases. The work done in this process is nothing but the change in the potential energies of final to initial. A sample case is taken and derived the equation for the work done as shown below.

Kinetic energy

Everybody revolving around the earth in a specific orbit will have some velocity and that velocity is called orbital velocity. As the body has orbital velocity, it consists of kinetic energy. We can measure the kinetic energy of the body using he conventional kinetic energy.

It could be noticed that the kinetic energy is positive and its magnitude is double the value of gravitational potential energy.

Total energy

We can find the total mechanical energy of a body as the sum of potential and kinetic energies. By adding both the values, it can be found that the total energy of the body is also has a magnitude of potential energy.




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Gravitation Complete Lesson

Gravitational force exits between every two bodies of the universe and it is directly proportional to the product of masses of the bodies and is inversely proportional to the square of distance of separation. This a universal law called Newton's law of gravitation. According to this law, the force is always attractive and it is independent of the medium existing between the two bodies. Thus the proportionality constant is constant over the entire universe.

The gravitational force between any two bodies is independent of the presence of the third body between them. When multiple bodies are present around any body, we can find the resultant force using the vector laws of addition. 

We also know that gravitation varies with height, depth and location of the body on the earth. We have derived mathematical equations for each of the case in the respective post.

The body on the surface of the earth possess both potential and kinetic energy and it can be measured from the center of the earth as a reference.

The body rotating in a specific orbit will have some velocity and it is called orbital velocity. The minimum velocity required to be give n to a body so that it can escape from the influence of the earth is called escape velocity and equations are derived for each of them in the following video lessons as mentioned below.

Gravitational Force of Attraction and Newton's Law




Orbital Velocity and Escape Velocity

Gravitation Problems with Solutions One

Gravitation Problems with Solutions Two


Gravitation Problems with Solutions Three

Gravitation Problems with Solutions Four


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