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An artificial satellite is moving in a circular orbit around the earth with a speed equal tohalf the magnitude of escape velocity from the earth.(a)Determine the heightof the satellite above the earth.s surface. (b) if the satellite is stopped suddenly in its orbit and allowed to fall freely onto the earth, find the speed with which it hits the surface of the earth.[g=9.8 m s^(-2) and R_E=6400 km] |
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Answer» Solution :(a) We KNOW that for SATELLITE motion `v_0=SQRT((GM)/r) =R sqrt(G/(R+h))` [as `g=(GM)/R^2` and r=R+h] In this problem `v_0=1/2v_e=1/2 sqrt(2gR)` [as `v_e=sqrt(2gR)` ] So `(R^2g)/(R+h)=1/2gR`, i.e., 2R=h+R or h=R=6400 km (b) By conservation of ME `0+[-(GMm)/r]=1/2mv^2+[-(GMm)/R]` or `v^2=2GM[1/R-1/(2R)]` [as r=R+h =R+R=2R] `v=sqrt((GM)/R)=sqrt(gR)=sqrt(10xx6.4xx10^6)`= 8 km/s |
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