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Naddika [18.5K]
3 years ago
11

Gravitational notes of physics ​

Physics
1 answer:
Pachacha [2.7K]3 years ago
7 0

Answer:

Every object in the universe attracts other object by a force of attraction, called gravitation, which is directly proportional to the product of masses of the objects and inversely proportional to the square of distance between them. This is called Law of Gravitation or Universal Law of Gravitation.

Let masses (M) and (m) of two objects are distance (d) apart. Let F be the attractional force between two masses.

Importance of The Universal Law of Gravitation

It binds us to the earth.

It is responsible for the motion of the moon around the earth.

It is responsible for the motion of planets around the Sun.

Gravitational force of moon causes tides in seas on earth.

Free Fall

When an object falls from any height under the influence of gravitational force only, it is known as free fall.

Acceleration Due to Gravity

When an object falls towards the earth there is a change in its acceleration due to the gravitational force of the earth. So this acceleration is called acceleration due to gravity.

The acceleration due to gravity is denoted by g.

The unit of g is same as the unit of acceleration, i.e., ms−2

Mathematical Expression for g

From the second law of motion, force is the product of mass and acceleration.

F = ma

For free fall, acceleration is replaced by acceleration due to gravity.

Therefore, force becomes:

F = mg ….(i)

But from Universal Law of Gravitation,

Factors Affecting the Value of g

As the radius of the earth increases from the poles to the equator, the value of g becomes greater at the poles than at the equator.

As we go at large heights, value of g decreases.

To Calculate the Value of g

Value of universal gravitational constant, G = 6.7 × 10–11 N m2/ kg2,

Mass of the earth, M = 6 × 1024 kg, and

Radius of the earth, R = 6.4 × 106 m

Putting all these values in equation (iii), we get:

Thus, the value of acceleration due to gravity of the earth, g = 9.8 m/s2.

Difference between Gravitation Constant (G) and Gravitational Acceleration (g)

S. No.

Gravitation Constant (G)

Gravitational acceleration (g)

1.

Its value is 6.67×10-11Nm2/kg2.

Its value is 9.8 m/s2.

2.

It is a scalar quantity.

It is a vactor quantity.

3.

Its value remains constant always and everywhere.

Its value varies at various places.

4.

Its unit is Nm2/kg2.

Its unit is m/s2.

Motion of Objects Under the Influence of Gravitational Force of the Earth

Let an object is falling towards earth with initial velocity u. Let its velocity, under the effect of gravitational acceleration g, changes to v after covering the height h in time t.

Then the three equations of motion can be represented as:

Velocity (v) after t seconds, v = u + ght

Height covered in t seconds, h = ut + ½gt2

Relation between v and u excluding t, v2 = u2 + 2gh

The value of g is taken as positive in case of the object is moving towards earth and taken as negative in case of the object is thrown in opposite direction of the earth.

Mass & weight

Mass (m)

The mass of a body is the quantity of matter contained in it.

Mass is a scalar quantity which has only magnitude but no direction.

Mass of a body always remains constant and does not change from place to place.

SI unit of mass is kilogram (kg).

Mass of a body can never be zero.

Weight (W)

The force with which an object is attracted towards the centre of the earth, is called the weight of the object.

Now, Force = m × a

But in case of earth, a = g

∴ F = m × g

But the force of attraction of earth on an object is called its weight (W).

∴ W = mg

As weight always acts vertically downwards, therefore, weight has both magnitude and direction and thus it is a vector quantity.

The weight of a body changes from place to place, depending on mass of object.

The SI unit of weight is Newton.

Weight of the object becomes zero if g is zero.

Weight of an Object on the Surface of Moon

Mass of an object is same on earth as well as on moon. But weight is different.

Weight of an object is given as,

Hence, weight of the object on the moon = (1/6) × its weight on the earth.

Try the following questions:

Q1. State the universal law of gravitation.

Q2. When we move from the poles to the equator, the value of g decreases. Why?

Q3. If two stones of 150 gm and 500 gm are dropped from a height, which stone will reach the surface of the earth first and why ?

Q4. Differentiate between weight and mass.

Q5. Why is the weight of an object on the moon 1/6th its weight on the earth??

You might be interested in
(Mass vs. Weight) HELP PLZ!!
Lina20 [59]

\huge \boxed { \sf{Answers}}

  • c. The weight of an object on the moon will be the same as its weight on Earth. It is false because the weight of an on the moon will be 1/6 th times its weight on Earth.
  • d. The weight of an object is its mass multiplied by the force of gravity. The statement is false because the formula of weight is mass × acceleration due to gravity, not force of gravity.
  • e. The mass and weight of an object are the same thing. The statement is false because mass means a body of matter. While weight of an object is its mass multiplied by the force of gravity.
  • f. The mass of an object is the force of gravity acting upon an object. It is false because it will be the weight of the object not mass.
  • So, the answers are c, d, e and f.

Hope you could understand.

If you have any query, feel free to ask.

7 0
3 years ago
A large sheet of charge has a uniform charge density of 9  μCm2. What is the electric field due to this charge at a point just
Alex73 [517]

Answer:

Explanation:

Surface charge density, σ = 9 μC/m² = 9 x 10^-6 C/m²

According to the Gauss theorem,

Electric field due to the sheet is given by

E = \frac {\sigma }{2\epsilon _{0}}

E = \frac{9\times 10^{-6}}{2\times 8.854\times 10^{-12}}

E = 5.08 x 10^5 N/C

7 0
3 years ago
A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo
zloy xaker [14]

Solution :

Given :

Mass attached to the spring = 4 kg

Mass dropped = 6 kg

Force constant = 100 N/m

Initial amplitude = 2 m

Therefore,

a). $v_{initial} = A w$

          $= 2 \times \sqrt{\frac{100}{4}}$

          = 10 m/s

Final velocity, v at equilibrium position, v = 5 m/s

Now, $\frac{1}{2}(4+4)5^2 = \frac{1}{2} kA'$

A' = amplitude = 1.4142 m

b). $T=2 \pi \sqrt{\frac{m}{k}}$

    m' = 2m

    Hence, $T'=\sqrt2 T$

c). $\frac{\frac{1}{2}(4+4)5^2 + \frac{1}{2}\times 4 \times 10^2}{\frac{1}{2} \times 4 \times 10^2}$

  $=\frac{1}{2}$

Therefore, factor $=\frac{1}{2}$

Thus, the energy will change half times as the result of the collision.

7 0
3 years ago
How much force is needed to accelerate a 1,500 kg car at a rate of 8 m/s2?
lesya [120]
Force = mass * acceleration = 1500kg * 8m/s²
3 0
3 years ago
A 782.10 kg car is brought from 7.60 m/s to 3.61 m/s over a time period of 4.23 seconds. What is the average force the car exper
alexandr402 [8]

Answer:

–735.17 N

The negative sign indicate that the force is acting in opposition direction to the car.

Explanation:

The following data were obtained from the question:

Mass (m) of car = 782.10 kg

Initial velocity (u) = 7.60 m/s

Final velocity (v) = 3.61 m/s

Time (t) = 4.23 s

Force (F) =?

Next, we shall determine the acceleration of the car. This can be obtained as follow:

Initial velocity (u) = 7.60 m/s

Final velocity (v) = 3.61 m/s

Time (t) = 4.23 s

Acceleration (a) =?

a = (v – u) / t

a = (3.61 – 7.60) / 4.23

a = –3.99 / 4.23

a = –0.94 m/s²

Finally, we shall determine the force experienced by the car as shown below:

Mass (m) of car = 782.10 kg

Acceleration (a) = –0.94 m/s²

Force (F) =?

F = ma

F = 782.10 × –0.94

F = –735.17 N

The negative sign indicate that the force is acting in opposition direction to the car.

4 0
3 years ago
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