Sunday 22 December 2013

GRAVITATION : Introduction, Detailed description, numericals, FAQs

Gravitations introduction, detailed explanation with equations, numericals, FAQsGRAVITATION : Introduction, Detailed description, numericals, FAQs


Introduction:


Gravity is the force by which a planet or other body draws objects toward its center. The force of gravity keeps all of the planets in orbit around the sun. Earth's gravity is what keeps you on the ground and what makes things fall.

Detailed Explanation with Equations:


Gravity is described by the law of universal gravitation, which states that every point mass attracts every other point mass by a force acting along the line intersecting both points. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them. The equation for the law of universal gravitation is given by:

F = G (m1m2/r²)

Where F is the force of gravity between the two objects, m1 and m2 are the masses of the two objects, r is the distance between the two objects, and G is the gravitational constant.

The gravitational constant, denoted by G, is a fundamental constant of nature that appears in the law of universal gravitation. Its value is approximately 6.674 x 10^-11 N·m²/kg².

The force of gravity is always attractive, which means that it pulls objects together. The magnitude of the force depends on the masses of the objects and the distance between them. If the masses of the objects are large, the force of gravity between them is also large. If the distance between the objects is large, the force of gravity between them is weaker.

Numericals:


Here are a few numerical examples to illustrate the use of the law of universal gravitation:

• Calculate the force of gravity between two objects with masses of 10 kg and 20 kg that are separated by a distance of 5 meters.

Solution:

F = G (m1m2/r²) F = 6.674 x 10^-11 N·m²/kg² (10 kg x 20 kg)/(5 m)² F = 2.004 x 10^-9 N

Therefore, the force of gravity between the two objects is 2.004 x 10^-9 N.

• A 1,000-kg satellite is in orbit around Earth at an altitude of 500 km. What is the force of gravity on the satellite?

Solution:

The distance between the satellite and the center of Earth is equal to the sum of the radius of Earth (6,371 km) and the altitude of the satellite (500 km).

r = 6,371 km + 500 km = 6,871 km = 6,871,000 meters

The mass of Earth is 5.97 x 10^24 kg.

F = G (m1m2/r²) F = 6.674 x 10^-11 N·m²/kg² (1,000 kg x 5.97 x 10^24 kg)/(6,871,000 m)² F = 8,869 N

Therefore, the force of gravity on the satellite is 8,869 N.

Problem: A planet of mass 6 x 10^24 kg has a radius of 6.4 x 10^6 m. A satellite of mass 1000 kg is orbiting around it at a height of 8000 km above the surface of the planet. Calculate: (a) The acceleration due to gravity on the surface of the planet. (b) The speed of the satellite in its orbit. (c) The time period of the satellite's orbit.

Solution: (a) The acceleration due to gravity on the surface of the planet can be calculated using the formula:

g = G M / R^2

where g is the acceleration due to gravity, G is the universal gravitational constant, M is the mass of the planet, and R is the radius of the planet.

Plugging in the given values, we get:

g = (6.67 x 10^-11 Nm^2/kg^2) x (6 x 10^24 kg) / (6.4 x 10^6 m)^2 g = 9.81 m/s^2

Therefore, the acceleration due to gravity on the surface of the planet is 9.81 m/s^2.

(b) The speed of the satellite in its orbit can be calculated using the formula:

v = √(G M / r)

where v is the speed of the satellite, G is the universal gravitational constant, M is the mass of the planet, and r is the distance of the satellite from the center of the planet.

The distance of the satellite from the center of the planet is:

r = R + h = (6.4 x 10^6 m) + (8000 km) = 6.4 x 10^6 m + 8 x 10^6 m = 1.44 x 10^7 m

Plugging in the given values, we get:

v = √[(6.67 x 10^-11 Nm^2/kg^2) x (6 x 10^24 kg) / (1.44 x 10^7 m)] v = 10,968 m/s (approx.)

Therefore, the speed of the satellite in its orbit is approximately 10,968 m/s.

(c) The time period of the satellite's orbit can be calculated using the formula:

T = 2πr / v

where T is the time period of the orbit, r is the distance of the satellite from the center of the planet, and v is the speed of the satellite.

Plugging in the values we have calculated, we get:

T = 2π x (1.44 x 10^7 m) / (10,968 m/s) T = 13,500 seconds (approx.)

Therefore, the time period of the satellite's orbit is approximately 13,500 seconds or 3.75 hours.

Problem: Two objects of masses 10 kg and 5 kg are placed at a distance of 2 m from each other. Calculate the gravitational force of attraction between them. Also, calculate the acceleration due to gravity experienced by each of the objects due to this force.

Solution: The gravitational force of attraction between the two objects can be calculated using the formula:

F = G (m1m2) / r^2

where F is the force of attraction, G is the universal gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between the centers of the two objects.

Plugging in the given values, we get:

F = (6.67 x 10^-11 Nm^2/kg^2) x (10 kg) x (5 kg) / (2 m)^2 F = 8.34 x 10^-10 N

Therefore, the gravitational force of attraction between the two objects is 8.34 x 10^-10 N.

The acceleration due to gravity experienced by each of the objects due to this force can be calculated using the formula:

a = F / m

where a is the acceleration due to gravity, F is the gravitational force of attraction, and m is the mass of the object.

For the 10 kg mass, the acceleration due to gravity is:

a = (8.34 x 10^-10 N) / (10 kg) a = 8.34 x 10^-11 m/s^2

For the 5 kg mass, the acceleration due to gravity is:

a = (8.34 x 10^-10 N) / (5 kg) a = 1.67 x 10^-9 m/s^2

Therefore, the acceleration due to gravity experienced by the 10 kg object is 8.34 x 10^-11 m/s^2 and the acceleration due to gravity experienced by the 5 kg object is 1.67 x 10^-9 m/s^2.

• Two objects of masses 5 kg and 10 kg are placed at a distance of 4 m from each other. Calculate the gravitational force of attraction between them.

Solution: Using the formula F = G (m1m2) / r^2, we get:

F = (6.67 x 10^-11 Nm^2/kg^2) x (5 kg) x (10 kg) / (4 m)^2 F = 8.34 x 10^-11 N

Therefore, the gravitational force of attraction between the two objects is 8.34 x 10^-11 N.

• A planet has a mass of 6 x 10^24 kg and a radius of 6.4 x 10^6 m. Calculate the acceleration due to gravity on the surface of the planet.

Solution: Using the formula g = G M / R^2, we get:

g = (6.67 x 10^-11 Nm^2/kg^2) x (6 x 10^24 kg) / (6.4 x 10^6 m)^2 g = 9.81 m/s^2

Therefore, the acceleration due to gravity on the surface of the planet is 9.81 m/s^2.

• A satellite of mass 1000 kg is in orbit around a planet with a mass of 6 x 10^24 kg and a radius of 6.4 x 10^6 m. The satellite is orbiting at a height of 8000 km above the surface of the planet. Calculate the speed of the satellite in its orbit.

Solution: The distance of the satellite from the center of the planet is:

r = R + h = (6.4 x 10^6 m) + (8000 km) = 6.4 x 10^6 m + 8 x 10^6 m = 1.44 x 10^7 m

Using the formula v = √(G M / r), we get:

v = √[(6.67 x 10^-11 Nm^2/kg^2) x (6 x 10^24 kg) / (1.44 x 10^7 m)] v = 10,968 m/s (approx.)

Therefore, the speed of the satellite in its orbit is approximately 10,968 m/s.

• A planet has a mass of 2 x 10^23 kg and a radius of 4000 km. Calculate the escape velocity from the surface of the planet.

Solution: Using the formula v = √(2GM/R), we get:

v = √[(2 x 6.67 x 10^-11 Nm^2/kg^2 x 2 x 10^23 kg) / (4000 km + 6.4 x 10^6 m)] v = 4.18 km/s (approx.)

Therefore, the escape velocity from the surface of the planet is approximately 4.18 km/s.

• Two objects of masses 2 kg and 4 kg are placed at a distance of 3 m from each other. Calculate the gravitational potential energy of the system.

Solution: Using the formula U = -G(m1m2)/r, we get:

U = -(6.67 x 10^-11 Nm^2/kg^2) x (2 kg) x (4 kg) / (3 m) U = -1.78 x 10^-10 J

Therefore, the gravitational potential energy of the system is -1.78 x 10^-10 J.

• A planet of mass 3 x 10^24 kg has a moon of mass 5 x 10^22 kg orbiting around it in a circular orbit at a distance of 4 x 10^5 km. Calculate the speed of the moon in its orbit.

Solution: Using the formula v = √(GM/r), we get:

v = √[(6.67 x 10^-11 Nm^2/kg^2 x 3 x 10^24 kg) / (4 x 10^5 km + 6.4 x 10^6 m)] v = 1024 m/s (approx.)

Therefore, the speed of the moon in its orbit is approximately 1024 m/s.

FAQs:


Q1. What is the difference between mass and weight?

Ans. Mass is a measure of the amount of matter in an object, while weight is a measure of the force of gravity on an object. Mass is measured in kilograms (kg), while weight is measured in newtons (N).

Q2. What is the acceleration due to gravity on Earth?

Ans. The acceleration due to gravity on Earth is approximately 9.8 m/s².

Q3. What is the difference between the gravitational force and the electrostatic force?

Ans. The gravitational force is the force of attraction between any two objects with mass, while the electrostatic force is the force of attraction or repulsion between two charged objects.

Q4. What is the gravitational field?

Ans. The gravitational field is the region around a mass where another mass would experience a force due to gravity. It is a vector field, meaning that at every point in space, there is a gravitational force acting in a certain direction.

Q5. What is escape velocity?

Ans. Escape velocity is the minimum velocity needed to escape the gravitational field of a planet or other celestial body. It is the velocity at which the kinetic energy of an object is equal to its potential energy due to gravity. The formula for escape velocity is given by:

v = sqrt(2GM/r)

Where v is the escape velocity, G is the gravitational constant, M is the mass of the planet or celestial body, and r is the distance from the center of the planet or celestial body.

Q6. What is the difference between gravitational potential energy and gravitational potential?

Ans. Gravitational potential energy is the energy an object possesses due to its position in a gravitational field, while gravitational potential is the potential energy per unit mass at a point in a gravitational field. Gravitational potential is a scalar quantity, while gravitational potential energy is a scalar quantity for a single object, but a vector quantity when considering the potential energy of a system of objects.

Q7. What is the difference between the weight of an object and the force of gravity acting on it?

Ans. The weight of an object is the force with which the object is pulled towards the center of the Earth due to gravity. It is equal to the mass of the object multiplied by the acceleration due to gravity. The force of gravity acting on an object is the force that the Earth exerts on the object due to its mass. It is equal to the product of the mass of the object and the gravitational field strength at the location of the object.

Q8. What is the significance of the gravitational constant?

Ans. The gravitational constant is a fundamental constant of nature that relates the strength of the gravitational force to the masses of the objects and the distance between them. It allows us to calculate the force of gravity between any two objects in the universe. The value of the gravitational constant is crucial in many areas of physics, including astronomy, cosmology, and particle physics.

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GRAVITATION


1.                              In The universe every mass attracts the other mass by a force, which is known as gravitational force.
2.                              Earth also attracts the other masses and the force of attraction of earth is said to be force of gravity.
Newton’s Law of Gravitation: -
This law is for the force of attraction between any two-point masses.
Consider two masses m1 & m2 , kept at a distance of  r  in between then the force acting between then will be   ‘F’ which is given as
<![endif]-->                             F                           F
               m1                                                                  m2

                                        r
1        The force of attraction is proportional to product of masses say
F  
<![endif]--> m1 m2
      2.   The force of attraction is inversely proportional to the square of distance between the masses. i.e.





                                           
3.    The gravitational force is always attractive force.






4.    The gravitational force will be along the line joining the masses.  So,
                          
       G  =  Gravitational Constant having same value in the complete universe.
       G  =    6.67 x 10 –11  Nm2     kg –2                                                  [G]  =  [M-1 L3 T-2]
       F   =    6.67 x 10 –11
Definition of unit mass or one kg   :-       
            If   m1 =  m2   =   1 kg ,       r  =   1 m
            F = 6.67 x 10-11  Nm2 kg-2

It two equal masses are kept at a distance of –1m apart and force acting between them is 6.67 x 10-11 N, then each mass will be unit mass of 1 kg.


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