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Take the mean distance of the moon and
the sun from the earth to be $0.4\times10 ^{6}$ km
and $150 \times10 ^{6}$ km respectively. Their
masses are $8 \times10 ^{22}$ kg and $2 \times10 ^{30}$ kg
respectively. The radius of the earth is
6400 km. Let $\Delta F_{1}$ be the difference in the
forces exerted by the moon at the nearest
and farthest points on the earth and $\Delta F_{2}$ be the difference in the force exerted by
the sun at the nearest and farthest points
on the earth. Then, the number closest to $\frac{\Delta F_{1}}{\Delta F_{2}}$ is :

Option 1)
2

Option 2)
$10^{-2}$

Option 3)
0.6

Option 4)
6

Answers (2)

best_answer

As we learned

Newton's Law of Gravitation -

$F\; \alpha\; \frac{m_{1}m_{2}}{r^{2}}$

$F\; = \frac{G\, m_{1}\, m_{2}}{r^{2}}$

$F\rightarrow$ Force

$G\rightarrow$ Gravitalional constant

$m_1,m_2\rightarrow$ Masses

$r\rightarrow$ Distance between masses

- wherein

Force is along the line joining the two masses

$f_{1} = \frac{-Gmme}{r_{1}2}, f_{2}=\frac{Gm_{e}m_{s}}{r_{2}^{2}}$

$\Delta f_{1}= \frac{-2GM_{e}m}{r_{1}^{3}} \Delta r1 and \Delta f_{2}=\frac{-2GM_{e}M_{s}}{r_{2}^{3}} \Delta r_{2}$

$\frac{\Delta f_{1}}{\Delta f_{2}} = \frac{M\Delta r}{r_{1}^{3}}\cdot \frac{Y_{2}^{3}}{M_{s}\Delta r_{2}}=\frac{m}{M_{s}}\cdot \frac{r_{2}^{3}}{r_{1}^{3}}\cdot \frac{\Delta r_{1}}{\Delta r_{2}}$

using $\Delta r_{1} = \Delta r_{2} = 2 Rearth m=8\times 10^{22}kg, M_{s}=2\times 10^{30}kg$

$r_{1} = 0.4\times 10^{6}km, r_{2}=150\times 10^{6}km$

we get $\frac{\Delta f_{1}}{\Delta f_{2}}=2$

Option 1)

2

Option 2)

$10^{-2}$

Option 3)

0.6

Option 4)

6

Posted by

Plabita

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