Here's a solution to a problem I thought of:

How fast does the Earth travel around the Sun?

Find: v
Given: Earth-Sun distance r(ES)=1.50E11 m, t= 1 y = 365 d
s=2(pi)r(ES)=2(pi)(1.50E11 m/1000 m/km)= 9.42E8 km
t=365 d x 24 h/d=8760 h
v=s/t=9.42E8 km / 8760 h = 1.08E5 km/h = 108 000 km/h

metric:s=9.42E8 km x 1000 m/km = 9.42E11 m
t=8760 h x 60 min/h x 60 s/min = 3.15E7 s
v=s/t=9.42E11 m / 3.15E7 s = 2.99E4 m/s = 29 919.9 m/s

Yep that agrees with the number I pulled from NASA :-)

=> 29.7 Km/sec which is around 1700 km/hr

Now do the next one the motion of the sun/earth about the galactic centre :-)

I agree with the value 29.7 km/s, but I don't think that is equal to 1700 km/h.

Here's how I figure it:

29.7 km/s x 60 s/min x 60 min/h = 106920 km/h

which is 63 times faster than your figure. What do you think?

It's not often I have a chance to disagree with Orac about anything finite but I have to say that I make it 106,920 km/h as well.

Yes someone multiplied by 60 rather than 3600 as I switching from meters to kilometers at same time didn't pick up the error.

You got me :-)