Numerical Problems

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

Distance between the centers = d = 0.5 m

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

Gravitational force = F = ?

        

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

 

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

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

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

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

Gravitational force = F = 0.006673 N
Gravitational constant = G = $6.673 \times 10^{-11} Nm^{2}kg^{-2}$
Distance between the masses = d = 1m
Mass = $m_{1}$ = $m_{2}$ =?
        
F = G $\frac{(m_{1} m_{2})}{d_{2}}$
F = G $\frac{( m \times m )}{d_{2}}$ $(Let \: m_{1} = m_{2}= m)$
F =$\frac{(m ^{2})}{d_{2}}$  

$m^{2}$ = $\frac{F \times d^{2}}{G}$
$m^{2}$ =$\frac{(0.006673 \times 〖(1) 〗^{2})}{( 6.673 \times 〖(10) 〗^{-11} )}$ =$\frac{(\frac{(6673}{1000000)}}{( 6.673 \times〖(10) 〗^{-11})}$ 
              =$\frac{(6.673 \times〖(10) 〗^{-3})}{( 6.673 \times〖(10) 〗^{-11})}$

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

Mass of Mars = $M_{m}$ = $6.42\times 1023$ kg

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

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

 

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

gm =$\frac{G M_{m}}{〖R〗^{2}_{m}}$     

 

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 $m^{-2}$