Section I: Part B - Short Answer Responses (1 Viewer)

[dolbinau]

I used Kepler's Law as well. made T the subject. substituted values for R and got numerical answers.. but then it wasn't exaclty a nice ratio so i rounded it a little bit and got 8:1. anyone do that? i hope i m not so confused that i have no idea what i'm talking about.

I got that answer using the same method...I think we're right. (well hope lol )

 

[adnan91]

OK if you did this: ORIGINAL PERIOD : T1

NEW PERIOD : 8T1

OR T2 = 8TI u r right

 

[gonutsmate]

For that keplers one, i think the r cubed comes into play.

 

[MacAus91]

i had
r^3/T^2 = (4r)^3/xT^2, where x is the ratio of periods
therefore, r^3/T^2 = 64r^3/xT^2
r^3/T^2 = 64r^3/(64T)^2
so x = square root of 64
=8

therefore the period is 8 times greater
ratio of periods = 8:1

 

[FelixRulz]

dang forgot to square root...
so ended up with 64:1