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Gravitation question

2018 · 15 Apr · Shift 1 · Q64
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  5. /2018 · 15 Apr · Shift 1 · Q64

Gravitation question

2018 · 15 Apr · Shift 1 · Q64

JEE MainPhysicsGravitationMCQ+4 / −1
Take the mean distance of the moon and the sun from the earth to be 0.4×1060.4 \times {10^6}0.4×106 km and 150×106150 \times {10^6}150×106 km respectively. Their masses are 8×10228 \times {10^{22}}8×1022 kg and 2×10302 \times {10^{30}}2×1030 kg respectively. The radius of the earth is 640064006400 km. Let ΔF1\Delta {F_1}ΔF1​ be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and ΔF2\Delta {F_2}ΔF2​ be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest to ΔF1ΔF2{{\Delta {F_1}} \over {\Delta {F_2}}}ΔF2​ΔF1​​ is :
  1. A
    222
  2. B
    10−2{10^{ - 2}}10−2
  3. C
    0.60.60.6
  4. D
    666
View written solutionFree

Correct answer: A

  1. Tidal-force type difference in force

The force due to a body of mass MMM on a test mass mmm at distance rrr is

F=GMmr2.F=\frac{GMm}{r^2}.F=r2GMm​.

If we compare the force at the nearest and farthest points on Earth relative to that body, the distances are approximately:

  • nearest point: r−REr-R_Er−RE​
  • farthest point: r+REr+R_Er+RE​

where RER_ERE​ is Earth's radius.

So,

ΔF=GMm(1(r−RE)2−1(r+RE)2).\Delta F = GMm\left(\frac{1}{(r-R_E)^2}-\frac{1}{(r+R_E)^2}\right).ΔF=GMm((r−RE​)21​−(r+RE​)21​).

Using

= \frac{(r+R)^2-(r-R)^2}{(r^2-R^2)^2} = \frac{4rR}{(r^2-R^2)^2},$$ and since $R_E \ll r$, we take $(r^2-R_E^2)^2 \approx r^4$. Hence, $$\Delta F \approx GMm\frac{4rR_E}{r^4}=\frac{4GMmR_E}{r^3}.$$ Thus, $$\Delta F \propto \frac{M}{r^3}.$$ Therefore, $$\frac{\Delta F_1}{\Delta F_2} = \frac{M_{\text{moon}}/r_{\text{moon}}^3}{M_{\text{sun}}/r_{\text{sun}}^3} =\frac{M_{\text{moon}}}{M_{\text{sun}}}\left(\frac{r_{\text{sun}}}{r_{\text{moon}}}\right)^3.$$ --- 2. **Substitute the given values** For the moon: $$M_m=8\times 10^{22}\ \text{kg}, \qquad r_m=0.4\times 10^6\ \text{km}$$ For the sun: $$M_s=2\times 10^{30}\ \text{kg}, \qquad r_s=150\times 10^6\ \text{km}$$ So, $$\frac{M_m}{M_s}=\frac{8\times 10^{22}}{2\times 10^{30}}=4\times 10^{-8}.$$ Also, $$\frac{r_s}{r_m}=\frac{150\times 10^6}{0.4\times 10^6}=\frac{150}{0.4}=375.$$ Hence, $$\left(\frac{r_s}{r_m}\right)^3=375^3.

Now,

3753=140625×375=52734375≈5.27×107.375^3=140625\times 375=52734375\approx 5.27\times 10^7.3753=140625×375=52734375≈5.27×107.

Therefore,

ΔF1ΔF2=4×10−8×5.2734375×107.\frac{\Delta F_1}{\Delta F_2}=4\times 10^{-8}\times 5.2734375\times 10^7.ΔF2​ΔF1​​=4×10−8×5.2734375×107.

=2.109375≈2.1.= 2.109375 \approx 2.1.=2.109375≈2.1.

  1. Closest option

The value closest to 2.12.12.1 is

2.\boxed{2}.2​.

So the correct option is A.

  1. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They agree.

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