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

g_m = G M_m/R²_m

g_h = GM_e/(R+h)²

g = GM_e/R

v_o = √GM_e/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 = m₂ = 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 = m₂ =?

F = Gm₁m₂/d²

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

m² = Fd²/G

m² = 0.006673 x (1)^-3 / 6.673*(10)^-11

Taking square root on both sides

√m² = √10⁸

m = 10⁴ 

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_m = 6.42 × 10²³ kg

Radius of Mars = R_m = 3370 km = 3370 × 1000 m = 3.37 × 10⁸ m

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

g_m = G Mm/R²m

OR      

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

g_m= 6.673 x 10^-11 x 6.42 x 10²³ / 11.357

g_m = 42.84/11.357

g_m = 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 = g_m = 1.62 ms^-2

Radius of the moon = R_m = 1740 km = 1740 × 1000 m = 1.74 × 10⁶ m

Mass of moon = M_m =?

g_m = GMm/R²m

OR

M_m =g_m R²_m / G  

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

M_m = 4.86 x 10¹² 10⁺¹¹ / 6.673

M_m = 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_e = 6.0 × 10²⁴ kg

Gravitational acceleration = g_h =?

g_h = GMe/(R+h)²

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

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

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

g_h = 6.673 × 10^-11 × 6.0 × 10¹⁰ = 40 × 10^-1 

g_h = 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 × 10⁴ x 10³ m = 4.87 x 10⁷ m

Gravitational acceleration = g =?

g = GM_e/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 10^1+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_e = 10000 km = 10000 × 1000 m = 10⁷ m

Mass of Earth = M_e =?

M_e = gR²/G

M_e = 4.0 x (10⁷)² / 6.673 x 10^-11

M_e = 0.599 × 10²⁵ kg

M_e = 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_e = 6.0 × 10²⁴ kg

Radius of Earth = R_e = 6.4 × 10⁶ m

Gravitational acceleration = g_h =  g = × 10 ms^-2 = 2.5 ms^-2

Altitude above Earth’s surface = h =?

g_h = GMe/(R+h)²

OR

(R + h)² = GMe/g_h

Taking square root on both sides

or     √(R+h)² = √GM_e/g_h

or        R + h = √G GM_e/g_h

or        h = √GM_e/g_h – 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 × 10⁶ m

As height is always taken as positive, therefore

h = 6.0 × 10⁶ m = 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 × 10⁶ m

Orbital velocity = v_o =?

v_o = √GM_e/R+h

v_o = √6.673*10^-11 * 6.0*10²⁴/6.4*10⁶ + 0.85*10⁶ =

v_o = √40.038*10¹³ /7.25*10⁶

v_o = √5.55 x 10⁷

 v_o  = 7.431 × 10³ 

v_o  = 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 × 10⁶ m

Orbital velocity = v_o =?

v_o = √GM_e/R+h

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

v_o  = √40.038*10¹³/48.4*10⁶

v_o   = √400.38*10¹²/48.4*10⁶

v_o  = √8.27*10⁶

v_o  = 2.876 × 10³ 

v_o  = 2876 ms^-1

Class 9 Physics Full Book Numerical Problems with Solution