JEE Mains · Physics · STD 11 - 7. gravitation
Take the mean distance of the moon and the sun from the earth to be \(0.4 \times 10^6\,km\) and \(150 \times 10^6\,km\) respectively. Their masses are \(8 \times 10^{22}\, kg\) and \(2 \times 10^{30}\, kg\) 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
- A \(2\)
- B \(6\)
- C \(10^{-2}\)
- D \(0.6\)
Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
As we know, Gravitational force of attraction, \(F = \frac{{GMm}}{{{R^2}}}\) \({F_1} = \frac{{G{M_e}m}}{{r_1^2}}\,and\,{F_2} = \frac{{G{m_e}{M_s}}}{{r_2^2}}\) \(\Delta {F_1} = \frac{{2G{M_e}m}}{{r_1^3}}\Delta {r_1}\,and\,\Delta {F_2} = \frac{{G{M_e}{M_s}}}{{r_2^3}}\Delta {r_2}\)…
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