1. Jupitor is 7.8 x 108 km from the sun and has a period
of 4.3 10% days. Find its tangential velocity in its
motion about the sun. [Ans. 4.75 x 104 km hrl]
Answers
tangential velocity in its motion about the sun is 4.75 × 10⁴ km/hr.
time period of Jupiter, T = 4.3 × 10³ earth days
= 4.3 × 10³ × 24 hrs
seperation between Jupiter and sun, r = 7.8 × 10^8 km
we know, time period is given as, T = 2π√(r³/GM) .........(1)
and tangential velocity of planet is given as v = √(GM/r) .........(2)
from equations (1) and (2),
we get, T = 2πr/v
or, v = 2πr/T
= 2π × (7.8 × 10^8 km)/(4.3 × 10³ × 24)
= (2 × 3.14 × 7.8 × 10^5)/(4.3 × 24)
= 0.47465 × 10^5 km/h
= 4.7465 × 10⁴ km/hr ≈ 4.75 × 10⁴ km/h
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time period of Jupiter, T = 4.3 × 10³ earth days
= 4.3 × 10³ × 24 hrs
seperation between Jupiter and sun, r = 7.8 × 10^8 km
we know, time period is given as, T = 2π√(r³/GM) .........(1)
and tangential velocity of planet is given as v = √(GM/r) .........(2)
from equations (1) and (2),
we get, T = 2πr/v
or, v = 2πr/T
= 2π × (7.8 × 10^8 km)/(4.3 × 10³ × 24)
= (2 × 3.14 × 7.8 × 10^5)/(4.3 × 24)
= 0.47465 × 10^5 km/h
= 4.7465 × 10⁴ km/hr ≈ 4.75 × 10⁴ km/h