kingwinner Posted June 5, 2006 Report Posted June 5, 2006 I have a couple more questions on this topic. I hope someone can help me. Thanks a lot! :( 1) A space shuttle ejects a 1.2x10^3 kg booster tank so that the tank is momentarily at rest, relative to Earth, at an altitude of 2.0x10^3 km. Neglect atmospheric effects. How much work is done on the booster tank by the force of gravity in returning it to Earth's surface? [The first thing I can think of is W=Fd(cos theta), but here the force of gravity is not constant so I can't use this formula. How can I find the work done by gravity, then?] 2a) Calculate the escape speed from the surface of the Sun: mass = 1.99x10^30 kg, radius = 6.96x10^8 m.2b) What speed would an object leaving Earth need to escape from our solar system? [For 2a), I got 6.18x10^5 m/s, but I don't know how to proceed to part :(] 3) Consider a geosynchronous satellite with an orbital period of 24h. What speed must the satellite reach during launch to attain the geosunchronous orbit? (Assume all fuel is burned in a short period. Neglect air resistance.) [On the ground, does the satellite have kinetic energy or is it at rest on the ground?] Quote
Jay-qu Posted June 5, 2006 Report Posted June 5, 2006 1. Intergration :( the equation W=Fd is derived from the area under (the intergral) of an F vs d graph, but the formular only works when F is constant, like you found out. For intergration this is no problem :( Quote
Tim_Lou Posted June 5, 2006 Report Posted June 5, 2006 2) calculate the gravitional potential (potential energy per unit of "test mass") and equate that 1/2 v^2. well for the "solar system" i guess you have to sum up the contributions from all the planets and the sun to calculate the total potential.... 3) from T, find r, using:[math]T^2/r^3={4\pi^2}/{Gm}[/math]from r, you can find the energy change and equate that to initial kinetic energy and find initial v. Quote
kingwinner Posted June 6, 2006 Author Report Posted June 6, 2006 Can somone please explain in greater detail? (especially for Q1) I still don't get how to do.:) Quote
Tim_Lou Posted June 6, 2006 Report Posted June 6, 2006 for 1 simply use the gravitional potential formula:[math]U=-Gm_1m_2/r[/math] (formula can be derive by integrating newton's law of gravity, with reference to r=infinity) find the difference in the potential energy, which would be the work done. Quote
kingwinner Posted June 6, 2006 Author Report Posted June 6, 2006 for 1 simply use the gravitional potential formula:[math]U=-Gm_1m_2/r[/math] (formula can be derive by integrating newton's law of gravity, with reference to r=infinity) find the difference in the potential energy, which would be the work done.Hi, Why only the change in potential energy? The speed changes too, so does the kinetic energy? Do I have to take the change in kinetic energy into consideration? Also, will the final answer of Q1 be a negative value? Quote
Jay-qu Posted June 6, 2006 Report Posted June 6, 2006 the kinetic energy changes the same amount as the gravitational (minus friction from atmosphere) Whether the work is positive or negative will depend on which way you deem positive. [math]W = \int(M_1M_2G/r^2)dr[/math] Quote
Tim_Lou Posted June 6, 2006 Report Posted June 6, 2006 yeah, jay is right, work=change in kinetic energy=change in gravitional energy. Quote
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