Ch:5-Gravitation (Numerical Problems)

 

Table of Contents

• Find the gravitational force of attraction between two spheres each of mass of 1000 kg
• The gravitational force between two identical lead spheres kept
• Find the acceleration due to gravity on the surface of Mars.
• The acceleration due to gravity on the surface of the moon
• Calculate the value of g at a height of 3600 km above the surface of the Earth.
• Find the value of g due to the Earth at geostationary satellite.
• Find the mass of the Earth.
• At what altitude the value of g would become one-fourth that on the surface of the Earth
• Find its orbital speed
• Find its orbital speed.

Find the gravitational force of attraction between two spheres each of mass of 1000 kg. The distance between the centers of the spheres is 0.5 m.
$(2.67\times 10^{-4}�)$

Difficulty: Hard

Solution:

Mass = ${�}_1={�}_2$= 1000 kg

Distance between the centers = d = 0.5 m

Gravitational constant = G = $6.673\times 10^{-11}�{�}^2�{�}^{-2}$

Gravitational force = F = ?

        

F =$\frac{(�{�}_1{�}_2)}{{�}_2}$ 

F = $6.673\times (10{)}^{-11}\times {\frac{(1000\times 1000)}{((0.5)}}^2)$

 

   =$\frac{6.673\times (10{)}^{-11}\times (10{)}^6}{0.25}$ 

   = $\frac{(6.673\times (10{)}^{-11}\times (10{)}^6)}{0.25}$

   =$\frac{(6.673\times (10{)}^{-5})}{0.25}$

   = $26.692\times (10{)}^{-5}$ = $2.67\times 10^{-4}�$

 

The gravitational force between two identical lead spheres kept at 1 m apart is 0.006673 N. Find their masses. (10, 000 kg each)

Difficulty: Hard

Solution:

Gravitational force = F = 0.006673 N
Gravitational constant = G = $6.673\times 10^{-11}�{�}^2�{�}^{-2}$
Distance between the masses = d = 1m
Mass = ${�}_1$ = ${�}_2$ =?
        
F = G $\frac{({�}_1{�}_2)}{{�}_2}$
F = G $\frac{(�\times �)}{{�}_2}$ $(���{�}_1={�}_2=�)$
F =$\frac{({�}^2)}{{�}_2}$  

${�}^2$ = $\frac{�\times {�}^2}{�}$
${�}^2$ =$\frac{(0.006673\times \text{〖}(1){\text{〗}}^2)}{(6.673\times \text{〖}(10){\text{〗}}^{-11})}$ =$\frac{(\frac{(6673}{1000000)}}{(6.673\times \text{〖}(10){\text{〗}}^{-11})}$ 
              =$\frac{(6.673\times \text{〖}(10){\text{〗}}^{-3})}{(6.673\times \text{〖}(10){\text{〗}}^{-11})}$

$\text{√}({�}^2)=10^8$ 
$�=10^4$ = 10000 kg each
Therefore, mass of each lead sphere is 10000 kg.

Find the acceleration due to gravity on the surface of Mars. The mass of Mars is $6.42\times 10^{23}$ kg and its radius is 3370 km.

Difficulty: Medium

Solution:

Mass of Mars = ${�}_{�}$ = $6.42\times 1023$ kg

Radius of Mars =${�}_{�}$ = 3370 km = $3370\times 1000$ m = $3.37\times 10^8�$

Acceleration due to gravity of the surface of Mars = ${�}_{�}$ =?

 

G =$\frac{6.673\ast 10^{-11}\ast 6.42\ast 10^{23}}{11.357}$ 

gm =$\frac{�{�}_{�}}{\text{〖}�{\text{〗}}_{�}^2}$     

 

or           gm = $6.673\times 10^{-11}\times$ $(6.42\times 10^{23})(3.37\times 10^6{)}^2$     

 

      =$\frac{(6.673\times 10^{(-11)}\times 6.42\times 10^{23})}{11.357}$ 

      =$\frac{42.84}{11.357}$ = 3.77 ${�}^{-2}$

The acceleration due to gravity on the surface of the moon is 1.62 

$�{�}^{-2}$. The radius of the Moon is 1740 km. Find the mass of the moon.

Difficulty: Medium

Solution:

Acceleration due to gravity = ${�}_{�}=1.62�{�}^{-2}$

Radius of the moon = ${�}_{�}$ = 1740 km = $1740\times 1000$ m = $1.74\times 10^6�$

Mass of moon = ${�}_{�}$ = ?

${�}_{�}$ = $�{�}_{�}\mathrm{/}\text{〖}�{\text{〗}}_{�}^2$    

 

or

${�}_{�}$ = $(\frac{��\times \text{〖}�{\text{〗}}_{�})}{�}$

${�}_{�}$ = $(\frac{1.62\times (1.74\times 10^6){)}^2}{(6.673\times 10^{-11})}$

 

= $(\frac{1.62\times 3\times 10^{12})}{(6.673\times 10^{-11})}$

${�}_{�}$ = $7.35\times 10^{22}$ kg

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

Difficulty: Medium

Solution:

Height = h = 3600 km = $3600\times 1000$ m = $3.6\times 10^6$ m
        Mass of Earth = ${�}_{�}$ = $6.0\times 10^{24}$ kg
        Gravitational acceleration = ${�}_{ℎ}$ =?

        ${�}_{ℎ}$ =$\frac{\text{〖}��{\text{〗}}_{�}}{(�+ℎ{)}^2}$
        ${�}_{ℎ}$ = $\frac{6.673\times 10^{-11}\times (6.0\times 10^{24})}{((6.4\times 10^6+3.6\times 10^6{)}^2)}$
              = $\frac{6.673\times 10^{-11}\times (6.0\times 10^{24})}{((10.0\times 10^6{)}^2)}$
              =$\frac{6.673\times 10^{-11}\times (6.0\times 10^{24})}{(100\times 10^{12})}$

= $6.673\times 10^{-11}\times 6.0\times 10^{10}$ = $40\times 10^{-1}$ = $4.0�{�}^{-2}$

Find the value of g due to the Earth at geostationary satellite. The radius of the geostationary orbit is 48700 km. ($0.17�{�}^{-2}$)

Difficulty: Medium

Solution:

Radius = R = $48700\times 1000$ m = $4.87\times 10^7$ m
Gravitational acceleration = g =?

        g =${\displaystyle \frac{(�{�}_{�})}{{�}^2}}$ 

        g =${\displaystyle \frac{6.673\times 10^{-11}\times (6.0\times 10^{24})}{((4.87\times 10^7{)}^2)}}$ 

           =${\displaystyle \frac{6.673\times 10^{-11}\times (6.0\times 10^{24})}{((23.717\times 10^{14}{)}^2)}}$ 

           =${\displaystyle \frac{(6.673\times 6.0\times 10^{-1})}{23.717}}$

           =${\displaystyle \frac{4.0038}{23.717}}$  

           = 0.17 $�{�}^{-2}$

The value of g is 4.0 

$�{�}^{-2}$ at a distance of 10000 km from the center of the Earth. Find the mass of the Earth. ($5.99\times 10^{24}$ kg)

Difficulty: Medium

Solution:

Gravitational acceleration = g = 4.0 $�{�}^{-2}$
        Radius of Earth = ${�}_{�}$ = 10000 km = $10000\times 1000$ m = $10^7$m
        Mass of Earth = ${�}_{�}$ =?

        ${�}_{�}$ =$\frac{(�{�}^2)}{�}$ 
        ${�}_{�}$ = $\frac{(4.0\times (10^7{)}^2)}{(6.673\times 10^{-11})}$ 

              =$\frac{(4.0\times 10^{14})}{(6.673\times 10^{-11})}$ 
              = $0.599\times 10^{25}$ kg = $5.99\times 10^{24}$ kg

At what altitude the value of g would become one-fourth that on the surface of the Earth? (One Earth’s radius)

Difficulty: Hard

Solution:

Mass of Earth = ${�}_{�}$ = $6.0\times 10^{24}$ kg

Radius of Earth = ${�}_{�}$ = $6.4\times 10^6$ m

Gravitational acceleration = gh =$\frac{1}{4}$ g =$\frac{1}{4}\times 10$ $�{�}^{-2}$ = $2.5�{�}^{-2}$

Altitude above Earth’s surface = h =?

gh = $\frac{\text{〖}��{\text{〗}}_{�}}{(�+ℎ{)}^2}$

            

or $(�+ℎ{)}^2$ = $\frac{\text{〖}��{\text{〗}}_{�}}{{�}_{ℎ}}$

Taking square root on both sides

    

$\text{√}16\ast 10^{13}{�}^2\text{–}6.4\ast 10^6$

or$\text{√}((�+ℎ{)}^2)$=$\frac{\text{√}(\text{〖}��{\text{〗}}_{�}}{{�}_{ℎ})}$ 

orR + h = $\frac{\text{√}(�\text{〖}��{\text{〗}}_{�}}{{�}_{ℎ})}$ 

orh =$\frac{\text{√}(�\text{〖}��{\text{〗}}_{�}}{{�}_{�}-�)}$ 

orh = $\text{√}((\frac{6.673\times 10^{-11}\times 6.0\times 10^{24})}{2.5)-6.4\times 10^6}$

   = $\text{√}((\frac{40.038\times 10^{13})}{2.5)-6.4\times 10^6}$

= $\text{√}(16\times 10^{13}{�}^2)-6.4\times 10^6$ = $\text{√}(0.16\times 10^{12})-6.4\times 10^6$

   = $0.4\times 10^6-6.4\times 10^6$

    = $-6.0\times 10^6$ m

            

As height is always taken as positive, therefore

h = $6.0\times 10^6$ m = One Earth’s radius

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

Difficulty: Medium

Solution:

Height = h = 850 km = $850\times 1000$ m = $0.85\times 10^6$ m

Orbital velocity = ${�}_{�}$=?${�}_{�}=\frac{\text{√}(\text{〖}��{\text{〗}}_{�}}{(�+ℎ))}$

 

${�}_{�}$ =$\frac{\text{√}((6.673\times 10^{-11}\times 6.0\times 10^{24})}{(6.4\times 10^6+0.85\times 10^6))}$

=$\frac{\text{√}((40.038\times 10^{13})}{(7.25\times 10^6))}$

      = $\text{√}(5.55\times 10^7)$ = $\text{√}(5.55\times 10^6$)

      = $7.431\times 10^3$ = $7431�{�}^{-1}$

A communication is launched at 42000 km above Earth. Find its orbital speed. 

$(2876�{�}^{-1})$

Difficulty: Easy

Solution:

Height = h = 42000 km = $42000\times 1000$ m = $42\times 10^6$m
        Orbital velocity = ${�}_{�}$ =?

$\text{√}8.27\ast 10^6$
        ${�}_{�}$=$\frac{\text{√}(\text{〖}��{\text{〗}}_{�}}{(�+ℎ))}$
        ${�}_{�}$ = $\frac{\text{√}((6.673\times 10^{-11}\times 6.0\times 10^{24})}{(6.4\times 10^6+42\times 10^6))}$ 
            =$\frac{\text{√}((40.038\times 10^{13})}{48.4\times 10^6))}$
              =$\frac{\text{√}((400.38\times 10^{12})}{(48.4\times 10^6))}$
              = $\text{√}(8.27\times 10^6)$
            = $2.876\times 10^3=2876�{�}^{-1}$