A planet in an orbit sweeps out an angle of 160° from march-may, when it is at at average distance of 140 million km from sun. If the planet sweeps out an angle of 10° from october-december, then the average distance from the sun is ..........
(1)56×10^5 km
(2)56×10^6 km
(3)56×10^7 km
(4)56×10^8 km
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"The correct answer to this question is Option (3) 56×10^7 km.
The average distance from the Sun will be 56×10^7 km.
R21Ø1= r22Ø2
(140 X 106)2 X 160 = r22 X 10
r22 = 140 X 106 X 4
r22 = 14 X10 X106 X4
r22 = 14 X4X 107
r22= 56 X 107. "
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