I did a question and the answer seems like it is too big for a satellite. Any help would be greatly appreciated! Calculate the total energy of a geosynchronous satellite (one that orbits over a fixed spot) with a mass of 1.5 x 10^3 kg, orbiting Earth at a height of 325km with an orbital speed of 5.0 x 10^3 m/s. m = 1.5 x 10^3 h = 325 000m v1 = 0 m/s v2 = 5.0 x 10^3 m/s Ep = mgh = 1.5 x 10^3(9.8)(325 000) = 4.8 x 10^9 J Ek = Ekfinal - Ekinitial = 1/2mv^2 - 1/2mv^2 = 1/2(1.5 x 10^3)(5.0 x 10^3)^2 - 1/2(1.5 x 10^3)(0)^2 = 1.9 x 10^10 Et = Ek + Ep = 1.9 x 10^10 + 4.8 x 10^9 = 1.4 x 10^10 J Thank you!
Hi, Actually, that's not so much. In fact, real geosynchronous orbits are at an altitude closer to 36,000 km above the Earth's surface, so the energies involved would be much higher. Also, Careful with this formula at high altitudes. At 325 km the Earth's gravitational pull is about 90% what it is at sea level, and ΔE<sub>p</sub> = mgh only works if g is nearly constant. A more accurate formula is E<sub>p</sub> = -GmM/r, (where r is the distance from the Earth's centre).
I see but I have never came across that formula, yet. Would it be absolutely wrong if I continue to use E<sub>p</sub> = mgh since it is the only formula that is in my book?
You'd lose some accuracy. In the case of your problem, the error in potential energy would be within 10%. Since most of the total energy is kinetic energy anyway, I suppose it wouldn't be the end of the world. Still, though, it's good to keep in mind that the formulas you're using have limits in their applicability.
Satellites do have a lot of kinetic and potential energy. Otherwise you wouldn't need big rockets (with lots of chemical energy) to get them into orbit. -Dale
