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A body is projected vertically upwards from the surface of the earth with a velocity sufficient to carry it to infinity . The time taken by it to reach a height of three times the radius of the earth is ( acceleration due to gravity = 9.8 ms^(-2) andradius of the earth = 6400 km) |
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Answer» 44.44 min `(TE)_("at weight")=(TE)_("at surface")` `(1)/(2)mv^(2)+(-(GMm)/(r))=(1)/(2)mv_(e)^(2)+(-(GMm)/(R ))` where `v_(e)` = escape velocity. `v^(2)=v_(e)^(2)+{(2GM)/(R)v[(R)/(r)-sqrt(2gR)g]=(GM)/R^(2)}` `V^(2)=2gR+2gR((R)/(r)-1)` `v=sqrt((2gR^(2))/(r)), So, v = (dr)/(dt)= Rsqrt((2g)/(r)) ""cdots(i)` `= R sqrt((2g)/(r))` Intergrating Eq. (i) we GET `int_(0)^(t)dt=(1)/(Rsqrt(2g))int_(R)^(R+h)r^(1//2).dr` `implies t=(2)/(3)(1)/(Rsqrt2g)[(R+h)^(3//2)-R^(3//2)]` `t=(1)/(3)sqrt((2R)/(g)[(1+(h)/(R))^(3//2)-1])` According to equestion h = 3R `t=(1)/(3)sqrt((2R)/(g)[(1+(3R)/(R))^(3//2)-1])` `t=(1)/(3)sqrt((2R)/(g))xx7` PUTTING all values we get `t=(1)/(3)((2xx6400xx10^(3))/(9.8))xx7` `[because R = 6400xx10^(3)]` `t=(80)/(3)XX 14.2xx7s` t=2666.65s t= 44.44min |
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