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gravitaion notes pdf maharashtra board
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1. study the entries in the following table and rewrite them putting the corrected items in a single row :gravitation

1 2 3
mass m/s2 zero at the center
weight kg measure of inertia
acceleration due to gravity Nm2/kg2 same in the entire universe
gravitational constant N depends on height

 Answer ::

 

1 2 3
mass N measure of inertia
weight kg zero at the center
acceleration due to gravity m/s2 depends on height
gravitational constant Nm2/kg2 same in entire universe

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?

Answer::

  Mass is the amount of matter present in an organism while weight of an object is mass along with gravitational force I. e. W=Mg.

 Mass of an object will be same at any point of the universe but its weight depends on the amount of gravity present on that planet thus mass of object on Mars will be same but weight will be different. {*this question can come for 2 marks in exam *}

 

b. What are (i) free fall, (ii) acceleration due to gravity (iii) escape velocity (iv) centripetal force ? 

Answer::
 

(i) Free Fall: 

Whenever an object moves under the influence of the force of gravity alone, it is said to be in Free Fall. During free fall initial velocity of an object is zero and force of air also acts on an object. Thus real free fall is possible only in a vacuum because there is no air.

(ii) Acceleration due to gravity:

The earth exerts gravitational force on objects near it. According to Newton’s second law of motion, a force acting on a body results in its acceleration. Thus, the gravitational force due to the earth on a body results in its acceleration. This is called acceleration due to gravity and is denoted by ‘g’. 

(iii) Escape velocity: 

The minimum value of the initial velocity at which an object to escape from the gravitational pull of the earth and never comes back to the earth is called the escape velocity.

Formula \(V_{esc}=\sqrt { \tfrac{2GM}{R}}\)

(iv) Centripetal force: 

The force that acts on an object to keep it moving along the circular path with constant speed is called the centripetal force. 

Its formula is \(F_c=\tfrac{mv^2}{r}\)

c. Write the three laws given by Kepler. How did they help Newton to arrive at the inverse square law of gravity?

Answer::
Kepler's first law:The orbit of a planet is an ellipse with the Sun at one of the foci.Kepler's Second law:The line joining the planet and the Sun sweeps equal areas in equal intervals of timeKepler's Third law :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.d. A stone thrown vertically upwards with initial velocity u reaches a height ‘h’ before coming down. Show that the time taken to go up is same as the time taken to come down.

Let us consider an object which is projected vertically upwards with initial velocity  u which reaches a maximum height h.
Acceleration due to gravity=-g
Equation of Motion for a body projected thrown upwards :
V=u-gt----------(1)
h=ut-1/2gt² -----------(2)
v²-u²=-2gh --------(3)


Equations of motion for freely falling body :
for free fall : 
Initial velocity=u=0
g=g
V=gt----------(4)
h=\(1\over2\)gt²-------(5)
v²=2gh------(6)

Time of Ascent is the time taken by body thrown up to reach maximum height h 
At maximum height , V=0

Equation (1) turns to u=gt1
\(t_1=\tfrac{u}{g} \) ----------(7)
Maximm height h=\(u²\over2g\) ----------(8)

Time of descent : After reaching maximum height , the body begins to travel downward like free fall
so equation (5) h=\(1\over2\)gt₂²
t₂²=\(2h\over g\)
\(t₂=\sqrt{\tfrac{2h}{g}}\)
but from equation (9)
t₂=\( \sqrt{\tfrac{2xu²}{2g²}}\)
t₂=u/g -----------------equation (10)
∴t₁=t₂
The time ascent is equal to time of descent in case of bodies moving under gravity

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? 

Answer:: Let the mass of the heavy object be m.
Thus, the weight of the object or the pull of the floor on the object is W=mg
Now, if g becomes twice, then the weight of the object or the pull of the floor on the object also becomes twice
i.e. W ′ =2mg=2W Thus, because of doubling of the pull on the object due to the floor,
it will become two times more difficult to pull it along the floor.


3. Explain why the value of g is zero at the centre of the earth.

Acceleration due to the earth's gravity is zero at the centre of the Earth because at that point the mass of the earth is equally distributed in all directions, so pulling equally in all directions for a net zero pull. As the distance from the centre decreases, the acceleration due to gravity also decreases.

Hence , the value of g is zero at centre .
 
 
 
 

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 \(\sqrt8\) T.

 
 

Thus, when the period of revolution of the planet at a distance R from a star is T, then from kepler's  third law of planetry motion, we have

\(T^2\propto R^3 \space\space\space\space.........(1)\)

now , when the distance of the planet from the star is 2R, then its period of revolution becomes

\(T_1^2\propto(2R)^3\)

\(T_1^2\propto8R^3 \)    ........(2)

Dividing equation 2 by 1

\(\tfrac{T_1^2}{T^2}=\tfrac{8R^3}{R^3}\)

\(T_1^2= 8 T^2\)

\(\therefore T_1=\sqrt8T\)

5. Solve the following examples.

a. An object takes 5 s to reach the ground from a height of 5 m on a planet. What is the value of g on the planet?

Answer::

Given : Time= 5s , Height=5m

To find : Gravitational acceleration 

Formula: \(S=ut+\tfrac{1}{2}gt^2\) 

steps: 

by formula

\(5=0\times t+\tfrac{1}{2}g(5)^2\)

\(5=\tfrac{1}{2}g\times25\)

\(\therefore g=\tfrac{2}{5}\)

\(\therefore g=0.4 m/s^2\)

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