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GingerBread27
Oct31-04, 10:21 AM
Calculate the torque required to accelerate the Earth in 3 days from rest to its present angular speed about its axis

I got that the Earth's mass is 5.98 X 10^24 kg and its radius is 6.38 X 10^6 m.

I used that the earth is solid sphere with a moment of inertia of:

I = 2/5*M*r^2 so then the Earth's moment of inertia would be: 9.74 X 10^37 kg*m^2.

To get torque i used T=(alpha)I, alpha being dw/dt, so I used (alpha)=(2pi)/(259200 s). 259200 seconds in three days.

I get the wrong torque, so what is wrong?

arildno
Oct31-04, 11:01 AM
I can't see where you're wrong.
It might be that your book has made the dreadful mistake of using degrees rather than radians, though.
That's about what I can offer you..

GingerBread27
Oct31-04, 11:13 AM
It's online hw so I know right away my answer is wrong, I get a torque of 2.36e33 Nm. I'm lost as to where I went wrong.

arildno
Oct31-04, 12:00 PM
Hmm.. I think I've found your mistake:
\omega_{e}=\frac{2\pi}{24h}=\frac{2\pi}{86400s}
3days=3*(86400s)
Hence,
\alpha=\frac{2\pi}{259200*86400s^{2}}
Hence, your answer should be reduced by a factor 86400.