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 = 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*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 = 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 *R²m/ G  

M_m = 1.62*(1.74*10⁶)²/6.673*10^-11

= 4.86*10¹²*10¹¹/6.673

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*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 = R_e = 10000 km = 10000 × 1000 m = 10⁷ m

Mass of Earth = M_e =?

M_e = gR²/G

M_e = 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 = 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*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

 

 

 

 

 

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 =

                              = √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 = v_o =?

v_o = √GM_e/R+h

v_o = √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