Asked by erin
How long would our year be if our Sun were two fifths its present mass and the radius of the Earth’s orbit were six times its present value?
Answers
Answered by
bobpursley
period^2= constant*r^3/M
T^2/1year=6^3/1* 1/(2/5)
T^2= 36*6*5/2
period= 6sqrt15 * 1year
check my thinking.
http://hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6
T^2/1year=6^3/1* 1/(2/5)
T^2= 36*6*5/2
period= 6sqrt15 * 1year
check my thinking.
http://hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6
Answered by
drwls
The gravity force would be less by a factor of (2/5)*(1/6)^2 = 1/90
The centripetal force would have to decrease my the same factor. That means
V^2/R becomes 1/90 of the former value.
Call V1 and R1 the original values and V2 and R2 the new values
V2^2/R2 = (1/90)V1^2/R1
V2^2 = (1/90)(R2/R1)*V1 = 1/15
V2 = 0.258 V1
The period of an orbit is proportional to R/V. It will increase by a factor
6/0.258 = 23.3
That is the new orbital period in years.
The centripetal force would have to decrease my the same factor. That means
V^2/R becomes 1/90 of the former value.
Call V1 and R1 the original values and V2 and R2 the new values
V2^2/R2 = (1/90)V1^2/R1
V2^2 = (1/90)(R2/R1)*V1 = 1/15
V2 = 0.258 V1
The period of an orbit is proportional to R/V. It will increase by a factor
6/0.258 = 23.3
That is the new orbital period in years.
Answered by
erin
dwls thank you very much
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