## Numerical Problems

#### 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 * (10)-11 * 1000*1000/(0.5)2

= 6.673 * (10)-11 * (10)6 /0.25

= 6.673 * (10)-5 /0.25

= 26.692 * (10)-5

= 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 = F*d2/G

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

Taking square root on both side

= 108

m = 104 = 10000 kg each

Therefore, mass of each lead sphere is 10000 kg.

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

Solution:        Mass of Mars = Mm = 6.42 × 1023 kg

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

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

gm = G Mm/R2m

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

= 6.673*10-11 * 6.42*1023 / 11.357

= 42.84/11.357 = 3.77 m-2

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

Solution:        Acceleration due to gravity = gm = 1.62 ms-2

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

Mass of moon = Mm = ?

gm = GMm/R2m

or           Mm =gm *R2m/ G

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

= 4.86*1012*1011/6.673

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 = Me = 6.0 × 1024 kg

Gravitational acceleration = gh =?

gh = GMe/(R+h)2

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

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

= 6.673 × 10-11 × 6.0*1024/(100*1012)

= 6.673 × 10-11 × 6.0 × 1010 = 40 × 10-1 = 4.0 ms-2

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

Solution:        Radius = R = 48700 × 1000 m = 4.87 × 107 m

Gravitational acceleration = g =?

g = GMe/R

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

= 6.673 × 10-11 × 6.0*1024/(23.17*1014)2

= 4.0038/23.717

= 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 = Re = 10000 km = 10000 × 1000 m = 107 m

Mass of Earth = Me =?

Me = gR2/G

Me = 4.0*(107)2 / 6.673*10-11

= 0.599 × 1025 kg = 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 = Me = 6.0 × 1024 kg

Radius of Earth = Re = 6.4 × 106 m

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)2 = GMe/gh

Taking square root on both sides

or     √(R+h)2 = √GMe/gh

or        R + h = √G GMe/gh

or        h = √GMe/gh – R

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

= √40.038*1013*/2.5 – 6.4*106

= √16*1013m2 – 6.4*106

= -6.0 × 106 m

As height is always taken as positive, therefore

h = 6.0 × 106 m = One Earth’s radius

#### (7431 ms-1)

Solution:        Height = h = 850 km = 850 × 1000 m = 0.85 × 106 m

Orbital velocity = vo =?

vo = √GMe/R+h

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

= √40.038*1013 /7.25*106

= √5.55*107

= 7.431 × 103 = 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 = vo =?

vo = √GMe/R+h

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

= √40.038*1013/48.4*106

= √400.38*1012/48.4*106

= √8.27*106

= 2.876 × 103 = 2876 ms-1