Table of Contents

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.

 

 

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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

 

 

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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

 

 

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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 = 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

 

 

 




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