Can ANy1 help?? on PAST PAPER 2005,Q16 (1 Viewer)

[backfire]

what is the answer for q16 a and b??? any1 out there who can help??? How is it done???
thanx

 

[Riviet]

For (a), use Kepler's Law of Periods: r³/T² = GM/4pi²
Rearranging this formula, T=sqrt(4pi²r³/GM)

T_Io=sqrt[(4pi².(421600000)³)/{(6.67x10^-11)(1.9x10²⁷)}]
= 1.53x10³² seconds.

We are given that the period of Ganymede is 2x the period of Europa, which is 2x the period of Io.

Therefore the period of Ganymede is 4x the period of Io:

= 4 x 1.53 x 10³²
= 6.12 x 10³² seconds.

Now we rearrange Kepler's law again to make r the subject:

r_(Ganymede)=(GMT²/4pi²)^1/3
=[(6.67x10^-11)(1.9x10²⁷)(6.12x10³²)/(4pi²)]^1/3
=1.06x10²⁷ m.

 

[helper]

In part a you were expected to be able to see and use the shortcut

r₁³/T₁²=r₂³/T₂²
r₁³=T₁²*r₂³/T₂²
r₁³=(2T)²*r₂³/T₂²
r₁³=4*r₂³
r₁³=4*421600000³
r₁=(4*421600000³)^1/3

Which makes the question a lot shorter and is why it was only worth 2 based on the marking guidelines
Also b requires either remebering an equation that needs an understanding of the definition of velocity or equating F_G=F_C

 

[passion89]

= 1.53x10³² seconds.

How did you get that? I'm doing it the same way and I keep getting roughly
152 643 seconds...
Am I missing something??