# Class 9 Physics Chapter 5 Numerical Problems

Class 9 Physics Chapter 5 numerical problems are according to the new syllabus of the Punjab Board. Chapter 5 is related to Gravitation in which you will study the force, motion, and field effects of gravitation.

## Class 9 Physics Chapter 5 Numerical Problems

The problems have been solved with easy-to-use methods using the following formulae:

F = Gm1m2/d2

gm = G Mm/R2m

gh = GMe/(R+h)2

g = GMe/R

v= √GMe/R+h

### Numerical Problems Chapter 5 – Gravitation

#### 5.1 Find the gravitational force of attraction between two spheres each of mass 1000 kg. The distance between the centers of the spheres is 0.5 m.             (2.67 × 10-4 N)

Solution:

Mass = m1 = m= 1000 kg

Distance between the centers = d = 0.5 m

Gravitational constant = G = 6.673 × 10-11 Nm2kg-2

Gravitational force = F =?

F = Gm1m2/d2

F = 6.673 x (10)-11 * 1000*1000/(0.5)2

F = 6.673 x (10)-11 * (10)6 /0.25

F = 6.673 x (10)-5 /0.25

F = 26.692 x (10)-5

F = 2.67 × 10-4 N

#### 5.2 The gravitational force between two identical lead spheres kept at 1 m apart is 0.006673 N. Find their masses.                                    (10,000 kg each)

Solution:

Gravitational force = F = 0.006673 N

Gravitational constant = G = 6.673 × 10-11 Nm2kg-2

Distance between the masses = d = 1m

Mass = m1 = m=?

F = Gm1m2/d2

F = m2/d2                     (Let m1 = m2 = m)

m= Fd2/G

m= 0.006673 x (1)-3 / 6.673*(10)-11

Taking square root on both sides

√m2 = √108

m = 10

m = 10000 kg each

Therefore, the mass of each lead sphere is 10000 kg.

#### 5.3 Find the acceleration due to gravity on the surface of Mars. The mass of Mars is 6.42 × 1023 kg and its radius is 3370 km. (3.77 ms-2)

Solution:

Mass of Mars = M= 6.42 × 1023 kg

Radius of Mars = Rm = 3370 km = 3370 × 1000 m = 3.37 × 10m

Acceleration due to gravity of the surface of Mars = gm =?

gm = G Mm/R2m

OR

gm = 6.673 × 10-11 × 6.42 x 1023 / (3.37 x 106)2

gm= 6.673 x 10-11 x 6.42 x 1023 / 11.357

gm = 42.84/11.357

gm = 3.77 ms-2

#### 5.4 The acceleration due to gravity on the surface of the moon is 1.62 ms-2. The radius of the Moon is 1740 km. Find the mass of the moon. (7.35 × 1022 kg)

Solution:

Acceleration due to gravity = gm = 1.62 ms-2

Radius of the moon = R= 1740 km = 1740 × 1000 m = 1.74 × 10m

Mass of moon = Mm =?

gm = GMm/R2m

OR

Mm =gm R2m / G

Mm = 1.62(1.74 x 106)2 / 6.673 x 10-11

Mm = 4.86 x 1012 10+11 / 6.673

Mm = 7.35 × 1022 kg

#### 5.5 Calculate the value of g at a height of 3600 km above the surface of the Earth.      (4.0 ms-2)

Solution:

Height = h = 3600 km = 3600 × 1000 m = 3.6 × 106 m

Mass of Earth = M= 6.0 × 1024 kg

Gravitational acceleration = gh =?

gh = GMe/(R+h)2

gh = 6.673 × 10-11 × 6.0 x 1024/(6.4 x 106+3.6 x 106)2

gh = 6.673 × 10-11 × 6.0 x 1024/(10.0 x 106)2

gh = 6.673 × 10-11 × 6.0 x 1024/(100 x 1012)

gh = 6.673 × 10-11 × 6.0 × 1010 = 40 × 10-1

gh = 4.0 ms-2

#### 5.6 Find the value of g due to the Earth at a geostationary satellite. The radius of the geostationary orbit is 48700 km.         (0.17 ms-2)

Solution:

Radius = R = 48700 × 1000 m = 4.87 × 10x 10m = 4.87 x 10m

Gravitational acceleration = g =?

g = GMe/R

g = 6.673 × 10-11 × 6.0 x 1024/(4.87 x 107)2

g = 6.673 × 10-11 × 6.0 x 1024 / 23.17 x 1014

g = 40.038 × 10-11+24 / 23.17 x 1014

g = 4.0038 x 101+13-14 / 23.717

g = 0.17 ms-2

#### 5.7 The value of g is 4.0 ms-2 at a distance of 10000 km from the center of the Earth. Find the mass of the Earth.                              (5.99 × 1024 kg)

Solution:

Gravitational acceleration = g = 4.0 ms-2

Radius of Earth = R= 10000 km = 10000 × 1000 m = 10m

Mass of Earth = M=?

Me = gR2/G

M= 4.0 x (107)2 / 6.673 x 10-11

Me = 0.599 × 1025 kg

Me = 5.99 × 1024 kg

#### 5.8 At what altitude the value of g would become one fourth than on the surface of the Earth?    (One Earth’s radius)

Solution:

Mass of Earth = M= 6.0 × 1024 kg

Radius of Earth = R= 6.4 × 10m

Gravitational acceleration = gh =  g = × 10 ms-2 = 2.5 ms-2

Altitude above Earth’s surface = h =?

gh = GMe/(R+h)2

OR

(R + h)= GMe/gh

Taking square root on both sides

or     √(R+h)= √GMe/gh

or        R + h = √G GMe/gh

or        h = √GMe/gh – R

or        h = √6.673 x 10-11 x 6.0 x 1024/2.5 – 6.4 x 106  – 6.4 × 106

h = √40.038*1013*/2.5 – 6.4*106

h = √16*1013m2 – 6.4*106

h  = -6.0 × 10m

As height is always taken as positive, therefore

h = 6.0 × 10m = One Earth’s radius

#### 5.9 A polar satellite is launched at 850 km above Earth. Find its orbital speed. (7431 ms-1)

Solution:

Height = h = 850 km = 850 × 1000 m = 0.85 × 10m

Orbital velocity = v=?

v= √GMe/R+h

v= √6.673*10-11 * 6.0*1024/6.4*10+ 0.85*10=

vo = √40.038*1013 /7.25*106

vo = √5.55 x 107

vo  = 7.431 × 10

vo  = 7431 ms-1

#### 5.10 A communication is launched at 42000 km above Earth. Find its orbital speed.       (2876 ms-1)

Solution:

Height = h = 42000 km = 42000 × 1000 m = 42 × 10m

Orbital velocity = v=?

v= √GMe/R+h

v= √6.673*10-11 * 6.0*1024/ 6.4*106+42*106

vo  = √40.038*1013/48.4*106

vo   = √400.38*1012/48.4*106

vo  = √8.27*106

vo  = 2.876 × 10

vo  = 2876 ms-1

Class 9 Physics Full Book Numerical Problems with Solution