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An earth satellite is revolving in a circular orbit of radius a withh a velocity `v_(0)`. A gun in the satellite is directly aimed toward earth. A bullet is fired from the gun with muzzle velocity `(v_(0))/2`. Find the ratio of distance of farthest and closest approach of bullet from centre of earth. (Assume that mass of the satellite is very-very large with respect to the mass of the bullet) |
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Answer» Correct Answer - 3 `v_(0)=sqrt((GM)/a)` `mv_(0)a=mvr` `1/2m(v_(0)^(2)+(v_(0)^(2))/4)-(GMm)/a=1/2mv^(2)-(GMm)/r` Solving `(r_(max))/(r_(min))=(2a)/(2a//3)=3` |
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