Gravitation [ in English ]

 Gravitation

Gravitation

Newton's law of gravitation :-

Every body in this universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between there centres. 

i e

Or

The constant G is called the universal gravitational constant

In S I  units the value of G is

In Vector form

Vector dirrected along AB

i e

dirrected along the line joining their centres

Or

Vector Form 

which is Newtons 3rd law of motion

=) gravitational force of attraction between two particles are equal in magnitude  but, opposite 

in sign

The centres of two masse

 So gravitational force is a central force.

Dimensional formula for G

Note the value of G does not depend upon the nature and size of the bodies.

* It does not depend upon The nature of medium between the two bodies.

Properties of gravitational Force

(1) gravitational force is always attractive

medium between two masses.

 (a) It is a central force i e  it acts along the line joining the centres of two bodies.

(4) It is the weakest force in nature 

(5) It is a conservative force I. e work done by it is path independent.

Gravity

The force of attraction exerted by the Earth on The body is called gravity or gravitational pul.l The acceleration produced in the motion of a body by the gravitational force of Earth is called acc. due to gravity and is denoted by g. The Value of g. by g on or hear the surface of earth is 9.8m/s². It is always directed vertically down towards the centre of the earth. 

Relation between g and -G

If the body of mass m lies The surface of the earth  or near its surface

Then

where M = mass of the Earth

          R = radius of the Earth

From this it follows that

g does not depend upon the mass m of a body.

Hence all bodies will fall on the earth  with same acceleration.

i.e all objects hollow or solid, big or small will strike the ground at the same time.

As g = (G / M) / R²

= G . (volume × density) / R²

= G   (4/3) * π  R³ * ρ / R²

= (4 /3 π GR ρ * R) / 3

Note: Mass of the Earth is 

M = 10²⁴ kg

and radius is

R = 6400 km

As we know density ρ = mass / volume

=> mass = volume × density

=> M = V * ρ

Variation of g

-------------------------

The value of g varies due

The value

to

(1) height (altitude)

2)Depth below the surface of south

(3) Shape of the Earth

(4) Rotation of the earth.

 

(1) the value of g at a height h above the surface of Earth

gh = GM/(R+h)२

where R= radius of Earth

If the body lies on the surface of the

Earth Then h=0

gh= GM/R२=g

Also gh= g ( 1-2h/R), h  R

(2) The value of g at the depth d below the surface of Earth is

gd = (1- d/R)

R

at the contre of the earth 

d=R

gd =0

Note the value of g is maximum on The surface of Earth it decreases with height and also with depth

(3) As we move from the equator towards the poles the value of g increases because Polar radius is less than equitorial radius.

g pole > g equates.

the weight of the body increases as it is taken from equator to the poles.

key points:-

em

(1)

g = GM/R२

(2)  gh = GM/(R+h)२

(3)  gh = g ( 1-2h/R)

(4) = gd= g ( 1- d/r)

At the centre of the Earth

 g = 0

Question 

If the distance between masses is doubled, the gravitational attraction between them

(a) is doubled 

(b) is reduced to quarter

Sol.

F = G m1m २/r२ 

putting &= 28 we get

F =  G m1m2/2r2

 

F = + F = F/4

gravitation force is red uced to Quarter

2 • Here =22 22

X F2 (2)242 24

: Ans if (b)

82 the distance between the ce ntres of moon and Earth is D. The mass of the Earth is 81 times the mase of the moon. At what distance from the centre of the Earth the gravitation-

=al force will be zero 4D 3 (a) (6) 20 (c) 40 (1) 90 10

Sall let mass of moon = M

1. Mass of Earth = 81M

Let P be a point at a distance x from the centre of the Earth where the gravitational force due to Earth and Moon is balanced (i.e., net force is zero).

Fe = Fm

(G * 81M * m) / x² = (G * M * m) / (D-x)²

where m is the mass of the object placed at P.

81 / x² = 1 / (D-x)²

9 / x = 1 / (D-x)

9D - 9x = x

10x = 9D

x = 9D / 10

Ans is (d).

Mass M is divided into two parts xM and (1-x)M. For a given separation, the value of x for which the gravitational attraction between the two pieces becomes maximum is

F = (G * xM * (1-x)M) / R²

F = (G * M² * (x - x²)) / R²

For maximum force, dF/dx = 0

(G * M² / R²) * (1 - 2x) = 0

1 - 2x = 0

x = 1/2

Ans is (a).

Two identical solid copper spheres of radius R are placed in contact with each other. The gravitational attraction between them is proportional to:

(a) R² (b) R³ (c) R⁴ (d) R⁻⁴

F = (G * m * m) / (R+R)²

F = (G * m²) / (4R²)

F = G(Mm)/R² = G (Vρ)(m) / R²

= G (4/3)πR³ρ * m / R²

= G * m * (4/3)π * ρ * R

Now squaring and continuing:

F = G * (mVρ)² / R²

= G * m² * (Vρ)² / R²

= G * m² * ((4/3)πR³ * ρ)² / R²

= G * m² * (16/9)π² * ρ² * R⁶ / R²

= (16/9)π² * G * m² * ρ² * R⁴

Therefore,

F ∝ R⁴

Ans is (C)

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Note:

As g = GM/R²

=> g = G(Vρ) / R²

= G * (4/3)π * R³ * ρ / R²

=> g = (4/3)πG * ρ * R

So,

g ∝ ρR

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Q1:

The radii of two planets are respectively R₁ and R₂, and their densities are respectively ρ₁ and ρ₂.

Find the ratio of acceleration due to gravity at their surfaces.

(a) g₁/g₂ = (R₁ * ρ₁) / (R₂ * ρ₂)

(b) g1 * g2 = R1 * P1 : R2 * P2

(c) g1 * g2 = R1 * P1 : R2 * P1

So, as g ∝ P R

g1 / g2 = P1 R1 / P2 R2

=> g1 : g2 = P1 R1 : P2 R2

Ans is (b)

Q6) If R is the radius of the earth and g the acc. due to gravity on the earth's surface, the mean density of the earth is

(a) πgR / 3gR  (b) 3πg / 4gR

(c) 3g / 4πRG  (d) πR - g / 12 - g

Sol: we have

g = - GM / R² = - G(PV) / R²

=> P = gR² / GV

=> P = gR² / G * 4/3 πR³

= 3g / 4πRG

Ans is (c)

Note: P = -M / V  => M = - PV

The mass and diameter of a planet have twice the value of the corresponding parameters of the Earth. 

Find the acceleration due to gravity on the surface of the planet.

(a) 9.8 m/s² (b) 4.9 m/s² (c) 98.0 m/s² (d) 19.6 m/s²

Solution:

On the surface of the Earth:

g = GM / R² = 9.8

On the surface of the planet:

g' = GM' / R'²

Since M' = 2M and R' = 2R,

g' = G(2M) / (2R)²

   = 2GM / 4R²

   = (1/2) * GM / R²

   = (1/2) * g

   = (1/2) * 9.8

   = 4.9 m/s²

Thus, the answer is (b).

Or alternatively,

g = GM / R²  =>  g ∝ M / R²

Thus, 

g' = M' / R'²

   = (2M) / (2R)²

   = 2M / 4R²

   = (1/2) * M / R²

   = (1/2) * g

   = 9.8 / 2

   = 4.9 m/s²

88. A body weighs 700 N on Earth. What would be its weight on a planet having 1/7th of Earth's mass and half Earth's radius?

(a) 400 N  (b) 300 N  (c) 200 N  (d) 100 N

Sol: Let m be the mass of the body then W = 700 N

=> mg = 700 N

We have M' = 1/7 M, R' = 1/2 R

Now g = GM/R^2, g' = G M' / R'^2

=> g' / g = (M' / M) x (R^2 / R'^2)

=> g' / g = (1/7) x (R^2 / (1/2 R)^2)

=> g' / g = (1/7) x (R^2 / (1/4 R^2))

=> g' / g = (1/7) x 4

=> g' = 4g/7

Weight of the body on the planet is

W' = mg' = m x (4g/7)

=> W' = (4/7) x 700

=> W' = 400 N

Ans is (a)

Q 9 The mass of the planet is 1/9th of the mass of the Earth and its radius is half that of the radius of the Earth. If a body weighs 450N on the Earth, what will be its weight on the planet?

a) 450N   b) 200N   c) 400N   d) 100N

Solution:

Let the mass of a body be M.

W = 450N on Earth

mg = 450N

We have M' = (1/9)M, R' = R/2

Now, g = M / R^2

g' = M' / (R')^2

g' = (1/9)M / (R/2)^2

g' = (1/9)M / (R^2/4)

g' = (1/9)M × 4 / R^2

g' = 4/9 g

g'/g = 4/9

g' = (4/9)g

Weight on the new planet:

W' = mg'

W' = m × (4/9)g

W' = (4/9) × 450

W' = 200N

Thus, the weight of the object on the planet is 200N.

Answer: (b)

Q10. If a planet consists of a satellite, whose mass and radius are both half that of Earth, then acc. due to gravity (g) at its surface should be —

(a) 29.4 m/s² (b) 19.6 m/s² (c) 9.8 m/s²

Sol: Let dash denote the parameter of a satellite

M' = M/2, R' = R/2

Now,

  g = GM / R²

⇒ g ∝ M / R²

⇒ g' / g = M' / M × R² / R'²

   = (1/2) × (2)²

   = (1/2) × 4 = 2

⇒ g' = 2g = 2 × 9.8 = 19.6

∴ g' = 19.6 m/s²

Ans is (b)

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Q11. Assume Earth to be a sphere of uniform density. What is the value of gravitational acc. in a mine 100 km below the earth’s surface? (given radius R = 6400 km)

(a) 9.1 m/s² (b) 9.65 m/s²

Sol: d = 100 km, R = 6400 km

Now, g_d = g (1 - d / R)

Gravitational Acceleration on a Planet

Q1. The mass and diameter of a planet have twice the value of the corresponding parameters of the Earth. Find the acceleration due to gravity on the surface of the planet.

(a) 9.8 m/s²

(b) 4.9 m/s²

(c) 9.80 m/s²

(d) 19.6 m/s²

Solution:

On the surface of the Earth,

g = GM / R² = 9.8

On the surface of the planet,

g' = GM' / R'² = G(2M) / (2R)² = 2GM / 4R² = (1/2)(GM / R²)

So,

g' = (1/2)g = (1/2) × 9.8 = 4.9 m/s²

Answer is (b)

Alternative Method:

g = GM / R²

 

 g => M / R²

So, g'/g = (M'/R'²) × (R²/M) = (2M × R²) / (M × (2R)²)

= (2R²) / (4R²) = 1/2

Therefore, g' = g / 2 = 9.8 / 2 = 4.9 m/s²

Q14. At what depth below the surface of the earth, is the value of g same as that at a height of 5 km?

(a) 10 km (b) 9.5 km (c) 5 km (d) 2.5 km

Sol: Here

        g_d = g(1 - d/R)

        g_h = g(1 - 2h/R)

Here  g_d = g_h

 ⇒ g(1 - d/R) = g(1 - 2h/R)

 ⇒ (1 - d/R) = (1 - 2h/R)

 ⇒ d/R = 2h/R

 ⇒ d = 2h

 ⇒ d = 2 × 5 = 10 km

∴ Ans is (a)

Q15. A body weighed 250 N on the surface assuming the earth to be a sphere of uniform mass density, how much would it weigh half way down to the...