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 = Gm_{1}m_{2}/d^{2}

g_{m} = G M_{m}/R^{2}_{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)

^{-4}N)

**Solution:**

Mass = m_{1} = m_{2 }= 1000 kg

Distance between the centers = d = 0.5 m

Gravitational constant = G = 6.673 × 10^{-11 }Nm^{2}kg^{-2}

^{ }Gravitational force = F =?

F = Gm_{1}m_{2}/d^{2}

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 }Nm^{2}kg^{-2}

Distance between the masses = d = 1m

Mass = m_{1} = m_{2 }=?

F = Gm_{1}m_{2}/d^{2}

F = m^{2}/d^{2 }(Let m_{1} = m_{2} = m)

m^{2 }= Fd^{2}/G

m^{2 }= 0.006673 x (1)^{-3} / 6.673*(10)^{-11}

Taking square root on both sides

√m^{2} = √10^{8}

m = 10^{4 }

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 × 10**^{23} kg and its radius is 3370 km. (3.77 ms^{-2})

^{23}kg and its radius is 3370 km.

**Solution:**

Mass of Mars = M_{m }= 6.42 × 10^{23} kg

Radius of Mars = R_{m} = 3370 km = 3370 × 1000 m = 3.37 × 10^{8 }m

Acceleration due to gravity of the surface of Mars = g_{m} =?

g_{m} = G Mm/R^{2}m

OR

g_{m} = 6.673 × 10^{-11} × 6.42 x 10^{23 }/ (3.37 x 10^{6})^{2}

g_{m}= 6.673 x 10^{-11} x 6.42 x 10^{23} / 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 × 10^{22 }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 × 10^{6 }m

Mass of moon = M_{m} =?

g_{m} = GMm/R^{2}m

OR

M_{m} =g_{m} R^{2}_{m} / G _{ }

M_{m} = 1.62(1.74 x 10^{6})^{2} / 6.673 x 10^{-11}

M_{m} = 4.86 x 10^{12} 10^{+11} / 6.673

M_{m} = 7.35 × 10^{22 }kg

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

^{-2})

**Solution:**

Height = h = 3600 km = 3600 × 1000 m = 3.6 × 10^{6} m

Mass of Earth = M_{e }= 6.0 × 10^{24 }kg

Gravitational acceleration = g_{h} =?

g_{h} = GMe/(R+h)^{2}

g_{h} = 6.673 × 10^{-11 }× 6.0 x 10^{24}/(6.4 x 10^{6}+3.6 x 10^{6})^{2}

^{ }g_{h} = 6.673 × 10^{-11 }× 6.0 x 10^{24}/(10.0 x 10^{6})^{2}

g_{h} = 6.673 × 10^{-11 }× 6.0 x 10^{24}/(100 x 10^{12})

g_{h}^{ }= 6.673 × 10^{-11 }× 6.0 × 10^{10 }= 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})

^{-2})

**Solution:**

Radius = R = 48700 × 1000 m = 4.87 × 10^{4 }x 10^{3 }m = 4.87 x 10^{7 }m

Gravitational acceleration = g =?

g = GM_{e}/R

g = 6.673 × 10^{-11 }× 6.0 x 10^{24}/(4.87 x 10^{7})^{2}

g = 6.673 × 10^{-11 }× 6.0 x 10^{24} / 23.17 x 10^{14}

g = 40.038 × 10^{-11+24} / 23.17 x 10^{14}

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 × 10^{24 }kg)

^{-2}at a distance of 10000 km from the center of the Earth. Find the mass of the Earth. (5.99 × 10

^{24 }kg)

**Solution: **

Gravitational acceleration = g = 4.0 ms^{-2}

Radius of Earth = R_{e }= 10000 km = 10000 × 1000 m = 10^{7 }m

Mass of Earth = M_{e }=?

M_{e} = gR^{2}/G

M_{e }= 4.0 x (10^{7})^{2} / 6.673 x 10^{-11}

M_{e} = 0.599 × 10^{25 }kg

M_{e} = 5.99 × 10^{24 }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 × 10^{24 }kg

Radius of Earth = R_{e }= 6.4 × 10^{6 }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 10^{24}/2.5 – 6.4 x 10^{6} – 6.4 × 10^{6}

h = √40.038*10^{13}*/2.5 – 6.4*10^{6}

h = √16*10^{13}m^{2} – 6.4*10^{6}

h^{ }= -6.0 × 10^{6 }m

As height is always taken as positive, therefore

h = 6.0 × 10^{6 }m = One Earth’s radius

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

^{-1})

**Solution:**

Height = h = 850 km = 850 × 1000 m = 0.85 × 10^{6 }m

Orbital velocity = v_{o }=?

v_{o }= √GM_{e}/R+h

v_{o }= √6.673*10^{-11 }* 6.0*10^{24}/6.4*10^{6 }+ 0.85*10^{6 }=

v_{o}^{ }= √40.038*10^{13 }/7.25*10^{6}

v_{o}^{ }= √5.55 x 10^{7}

^{ }v_{o}^{ } = 7.431 × 10^{3 }

v_{o}^{ } = 7431 ms^{-1}

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

^{-1})

**Solution:**

Height = h = 42000 km = 42000 × 1000 m = 42 × 10^{6 }m

Orbital velocity = v_{o }=?

v_{o }= √GM_{e}/R+h

v_{o }= √6.673*10^{-11 }* 6.0*10^{24}/ 6.4*10^{6}+42*10^{6}^{ }

v_{o}^{ } = √40.038*10^{13}/48.4*10^{6}

v_{o} = √400.38*10^{12}/48.4*10^{6}

v_{o} = √8.27*10^{6}

v_{o} = 2.876 × 10^{3 }

v_{o}^{ } = 2876 ms^{-1}

**Class 9 Physics Full Book Numerical Problems with Solution**