Points To Be Learn
1] Gravitation 2] Circular motion and centripetal force
3] Kepler’s laws 4] Newton’s universal law of gravitation
5] Acceleration due to the gravitational force of the Earth
6] Free fall 7] Escape velocity
Gravitation or gravity is a natural phenomenon by which all things in the universe which possess energy are brought towards one another, including stars, planets, galaxies, and even light and subatomic particles.
Gravitational Force: 1] This is an attractive force existing between any two objects 2] It is a universal force in which every object in the universe experiences this force due to the presence of other objects. 3] larger objects that exert this force on smaller objects, e.g. When you throw a stone up from the earth, it will fall due to the earth’s gravitational force. 4] Similarly, Moon is orbiting around the earth due to the gravitational force of the earth. Planets are orbiting around the Sun due to the gravitational force of the Sun.
According to Newton’s law of gravitation, the gravitational force is directly proportional to the product of the two masses and Inversely proportional to the square of the distance between them.
Force and Motion:
In physics, a force is any interaction that when unopposed, will change the motion of an object A force can cause an object with mass to change its velocity i.e., to accelerate or decelerate.
Circular motion:
It is a movement of an object along the circumference of a circle or rotation along a circular path to can be uniform, with constant angular rate of rotation and constant speed, or non uniform with a changing rate of rotation
Uniform circular motion :
It can be described as the motion of an object in a circle at a constant speed. As an object moves in a circle, it is constantly changing its direction. At all instances the object is moving tangentially to the circle.
Non – Uniform circular motion :
This type of motion is. defined as the motion of an object in which the object travels with varied speed and it does not cover equal distance in equal intervals of time, irrespective of the time interval duration
Centripetal force :
It is a force on an object directed towards the center of a circular path that keeps the object moving on the path. It’s value is based on three factors.
(i) the velocity of the object as it follows the circular path.
(ii) the object’s distance from the centre of the path.
(ii) the mass of the object.
Kepler’s laws of planetary motion
1) Kepler’s first law : The orbit of a planet is an ellipse withsun at one of the foci. 2) Kepler’s second law : The line joining the planet and the sun sweeps equal area in equal intervals of time. 3) Kepler’s third law : The square of its period of revolution around the sun is directly proportional to the cube of the mean distance of the planet from the sun.
Earth’s Gravitational Acceleration :
It is defined as the acceleration gained by an object due to force of gravity acting on it. It is represented by ‘g’. It is a vector quantity, i.e. it possesses both magnitudes as well as direction. It can be calculated by using the formula.
The value of ‘g’ depends on the location of the body.
Mass : Mass is defined as the amount of matter present in an object. The SI unit of mass is kg. Mass is a scalar quantity. It’s value is same everywhere.
Weight: The weight of an object is defined as the force with which the earth attracts the object towards itself.
Free fall :
It is defined as the motion of an object where gravity is the only force acting upon it. For a freely falling object, the velocity on reaching the earth and time taken for it can be calculated by using Newton’s equation of motion for free fall, the initial velocity v = 0 acceleration a = g. Thus we can write the equation of motion as follows.
V2= u’+ 2gs For object thrown upwards acceleration is –ve and g=-g
Gravitational Potential energy :
The energy possessed by an object because of its position in the gravitational field is called Gravitational Potential Energy. The formula for Gravitational Potential Energy on earth’s surface is expressed as V = mgh The formula for Gravitational potential energy at a height ‘h’ from the surface of the earth is expressed as follows,
Escape velocity :
Escape velocity is the minimum velocity required by a body to overcome the gravitational pull of the earth. Escape velocity is given by
The total energy of body revolving around the earth is the sum of kinetic energy and potential energy.
Total energy = kinetic energy+potential energy.
Question Answer : Exercise
Q.1.Study the entries in the following table and rewrite them by putting the connected items in a single row.
Q.2. Answer the following questions.
(a) What is the difference between mass and weight of an object. Will the mass and weight of an object on the earth be same as their values on Mars? Why? Ans. (1) Difference between mass and weight of an object. MASS:A] The mass of an object is the amount of matter present in a substance. 2] Mass is same everywhere in the Universe and is never zero. 3] It is a scalar quantity and its SI unit is kg. WEIGHT: 1] The weight of an object is the force with which the earth (or any other planet/ moon/star) attracts it. 2] It is directed towards the centre of the earth. 3] The weight of an object is different at different places on the earth. 4] It is zero at the earth’s centre. 5] It is a vector quantity and its SI unit is the newton (N).
B] 1]Mass of an object on earth will be the same as its mass on Mars. because mass is simply a measure of inertia. It is the amount of matter present in the object. But the weight of an object on earth will not be the same as its weight on Mars, because weight is a force exerted on a body by gravity. This is often expressed as
W=F= mg,
where m = mass of an object g = acceleration due to gravity of that planet. For Earth, g=9.8 m/s2 For Mars, g=3.7 m/s2
As the value of g for Earth and Mars are different, the weight of an object on them will be different. As the value of g for Earth and Mars are different, the weight of an object on them will be different. Weight of an object on Mars will be less than that on earth.
(b) What is (i) Free fall (ii) Acceleration due to gravity (iii) Escape Velocity (iv) Centripetal force. Ans. (I) Free fall – When the object moves under the influence of the force of gravity alone, it is said to be in a state of free fall. During free fall (i) Initial velocity u = 0. (ii) Force acting on object is only gravitational force. (iii) Acceleration produced is a=g=9.8 m/s2. (iv) Frictional forces and buoyant forces due to air or surroundings are absent. (v) Free fall can be observed only in a vacuum. (II) Acceleration due to gravity – The acceleration obtained by an object because of the gravitational force is called acceleration due to gravity. The acceleration due to gravity is represented by g. It has a value of 9.80665 m/s2. Its value is different at different places.
(iii) Escape Velocity – Escape velocity is the minimum speed needed for an object to escape from the gravitational influence of a body. Or The minimum velocity required for a satellite to become free from Earth’s gravitational influence is called as escape velocity. The escape velocity for earth is about 11.186 km/s at the surface.
(iv) Centripetal Force – It is a force on an object directed towards the center of a circular path that keeps the object Direction of on the path.
Its value is based on the three factors. (a) The velocity of the object as it follows the circular path. (b) The object’s distance from the centre of circular path. (c) The mass of the object. (c) Write the three laws given by Kepler. How did they help Newton to arrive at the inverse square law of gravity? Ans. Kepler’s law – Kepler gave three laws of planetary motion. Kepler’s Ist law (Law of orbit) – The orbit of a planet is an ellipse with the Sun at one of the foci. Figure shows the elliptical orbit F of a planet revolving around planet the sun has the position of the Sun indicated by S.
Kepler’s IInd law (Law of equal areas) – The line joining the planet and the Sun sweeps equal areas in equal intervals of time. AB and CD are distances covered by the planet in equal time i.e. after equal intervals of time, the positions of the planet starting from a and Care shown by B and D respectively. The straight lines AS and CS sweep equal area in equal intervals of time i.e., area ASB and CSD are equal.
Kepler’s IIIrd law (Law of Period) – The square of its period of revolution around the Sun is directly proportional to the cube of the mean distance of a planet from the sun. Thus, if r is the average distance of the planet from the Sun and T is its period of revolution then
.
(d) A stone thrown vertically upwards with initial velocity ‘u‘ reaches height ‘h‘ before coming down. Show that the time taken to go up is same as the time taken to come down. Ans. Time taken to go up: Let the time taken to go up bet, Here, the stone reaches the maximum height h as its velocity decreases to zero .: s=h, v=0 , a=-g Here, g is taken to be negative because it is a case of retardation. Now, we know v=u+at On putting the values in the above equation we get .
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Time taken to go down: Let the time taken to go down be t. Here we have u=0, s = h, a=g On putting these values in
(e) If the value of g suddenly becomes twice its value, it
will become two times more difficult to pull a heavy
object along the floor. Why? Ans. The difficulty in putting the object depends upon its weight. More is the weight more is the difficulty.
Hence let us calculate the weight of an object.
Case I :- Before changing ‘g’
Mass = m,
acceleration due to gravity = g
.: Weight =w, = mg ………………( i )
Case II :- After changing ‘g’
Mass = m,
acceleration due to gravity =2g
.: Weight = w2 = mx 2g = 2mg ……… (ii)
Here, we see w2 = 2w1…………………………. From (i) & (ii)
i.e. when g becomes twice, weight also becomes twice.
Hence, force required to pull becomes twice. Thus it becomes twice difficult to lift the object.
Q.3. Explain why the value of g is zero at the centre of the earth.
Ans. The general equation for acceleration due to gravityʻgʻis g = GM / r2
Assuming earth to be a perfect sphere, and considering it’s uniform density. Multiplying and dividing the RHS by volume ‘V’
Q.4. Let the period of revolution of a planet at a distance R from a star be T. Prove that if it was at a distance of 2R from the star, its period of revolution will be √8 T.
Where T = Time period of revolution, r = Distance of planet from star,
K = Constant.
Q.5. Solve the following examples. (a) An object takes 5s to reach the ground from a height of 5m on a planet. What is the value of g on the planet?
Given : t=5s, s=5m, u=0
To find : g =?
Soln: We have, s = ut + 1/2 at2
Very nice sir î and thank you so much for science points and notes .