gcse physics questionWatch
a satellite on a polar orbit has a time period of 90 mins.
the earth's gravitational field strength from that height is 10N/kg.
given that the radius of the earth is 6400 km, find the height of the satellite above the earth's surface?
you mean a=(v^2)/r...but how would u work out 'v'
where omega = 2 pi/T
T = 90 mins = 90 * 60 = 5400 seconds.
i think this question required some major algebraic manipulation...but unfortunately i didn't have enough time to finish it.
did u do edexcel for maths as well...what did u think of that calc exam?
the thing is, i don't actually know the radius of the satellite's orbit- i only know the earth's (6400 km)...so you're not able to use v=(2*pi*r)/T, or are you?
Can you not then use a = r(omega)^2, using a = 10, because the satellite is in freefall?
a = (6400 x 1000 + h)(2 pi/5400)^2.
Since when has this sort of stuff been in GCSE Physics anyway? I did my GCSEs in 2001 and we weren't required to know anything about circular motion/satellite orbits.
circular motion has always been there, but is only on the extension paper (1 hr). This is just an add-on to the core syllabus, which for physics, includes particles and communications. whether you do the ext paper depends on whether your school decides to do it or not.