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Q.

At its aphelion the planet mercury is $7 \times 10^{10} m$ from the sun & at its perihelion it is $4 .5 \times 10^{10} m$ from the sun. If the speed of mercury at aphelion is $3 {.6×10}^{4} m/s$ , then speed of it at perihelion is

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a

3 {.6×10}^{4} m/s

b

5 {.6×10}^{4} m/s

c

1 {.6×10}^{4} m/s

d

Mass of mercury must be known to compute the required value from given information

answer is B.

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Detailed Solution

According to L.C. Angular momentum

${m v}_{1} r_{1} = {m v}_{2} r_{2}$

$\frac{v_{1}}{v_{2}} = \frac{r_{2}}{r_{1}} \Rightarrow \frac{3.6 \times 10^{4}}{v_{2}} = \frac{4 .5 \times 10^{10}}{7 \times 10^{10}}$

$v_{2} = \frac{3.6 \times 7 \times 10^{4}}{4.5} = 5.6 \times 10^{4} m/s$

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At its aphelion the planet mercury is7×1010m from the sun & at its perihelion it is 4.5×1010m from the sun. If the speed of mercury at aphelion is3.6×104m/s , then speed of it at perihelion is