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
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?
3 answers
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.
dwls thank you very much