Chapter 5, Problem 41P

To determine

  $\text{9}{\text{.6×10}}^{\text{17}}\text{N}$

The total force on Earth due to Venus, Jupiter and Saturn, and the fraction of Sun’s force on the Earth is this total force.

Answer to Problem 41P

Solution: Total force on Earth due to Venus, Jupiter, and Saturn=.

This is $2.7\times {10}^{-5}$ times the force exerted by Sun alone.

Explanation of Solution

Given:

Mean Distance of Venus from Sun =108 million km

Mean Distance of Venus from Jupiter =778 million km

Mean Distance of Venus from Saturn=1430 million km

The mass of Venus is 0.185

The mass of Jupiter is 318

The mass of Saturn is 95.1

Formula Used:

Gravitational force between two masses m and M which are r distance apart can be obtained by:

  $F=\frac{GmM}{r^2}$

Where, G is the gravitational constant.

Calculation:

  $\begin{array}{c}{r}_{\begin{array}{l}Earth\\ Venus\end{array}}=(150-108)\times {10}^6\text{km}\\ =4.2\times {10}^{10}\text{m}\\ r_{\begin{array}{l}Earth\\ Jupiter\end{array}}=(778-150)\times {10}^6\text{km}\\ =6.28\times {10}^{11}\text{m}\\ r_{\begin{array}{l}Earth\\ Saturn\end{array}}=(1430-150)\times {10}^6\text{km}\\ =1.28\times {10}^{\text{12}}\text{m}\end{array}$

Jupiter and Saturn will exert a rightward force while Venus will exert a left-ward force take the night direction as positive

  $F=G\frac{M_{Earth}{M}_{Jupiter}}{r^2{}_{\begin{array}{l}Earth\\ Jupiter\end{array}}}+G\frac{M_{Earth}{M}_{saturn}}{r^2{}_{\begin{array}{l}Earth\\ saturn\end{array}}}-G\frac{M_{Earth}{M}_{Venus}}{r^2{}_{\begin{array}{l}Earth\\ Venus\end{array}}}$

  $=G{M}^2{}_{Earth}\left(\frac{\text{318}}{{\text{(6}{\text{.28×10}}^{\text{11}}\text{m)}}^{\text{2}}}\text{+}\frac{\text{95}\text{.1}}{{\left(\text{1}{\text{.28×10}}^{\text{12}}\text{m}\right)}^{\text{2}}}\text{-}\frac{\text{0}\text{.815}}{{\left(\text{4}{\text{.2×10}}^{\text{10}}\text{m}\right)}^{\text{2}}}\right)$

  $\begin{array}{l}\text{=}\left(\text{6}{\text{.67×10}}^{\text{-11}}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\text{(5}{\text{.97×10}}^{\text{24}}\text{kg)(4}{\text{.02×10}}^{\text{-22}}{\text{m}}^{\text{-2}}\text{)}\\ \text{=9}{\text{.56×10}}^{\text{17}}\text{N}\approx \text{9}{\text{.6×10}}^{\text{17}}\text{N}\end{array}$

  $F_{\begin{array}{l}Earth\\ Sun\end{array}}=G\frac{M_{Earth}{M}_{Sun}}{r^2{}_{\begin{array}{l}Earth\\ Sun\end{array}}}\left(\text{6}{\text{.67×10}}^{\text{-11}}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\frac{\left(\text{5}{\text{.97×10}}^{\text{24}}\text{kg}\right)\left(1.99\times {10}^{10}\text{kg}\right)}{{\left(\text{1}{\text{.50×10}}^{\text{11}}\text{m}\right)}^2}$

And so, the ratio is

  $\begin{array}{l}\frac{F_{\begin{array}{l}earth\\ planets\end{array}}}{F_{\begin{array}{l}earth\\ sun\end{array}}}=\frac{9.{\text{56×10}}^{\text{17}}\text{N}}{\text{3}{\text{.52×10}}^{\text{22}}\text{N}}\\ f=2.7\times {10}^{-5}\end{array}$

Conclusion:Total force on Earth due to Venus, Jupiter, and Saturn= $\text{9}{\text{.6×10}}^{\text{17}}\text{N}$

This is $2.7\times {10}^{-5}$ times the force exerted by Sun alone.