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The critical velocity of a satellite at height h from the surface of the earth is 5 km/s. The same for a satellite around another planet of double the radius and 4 times the mass of the earth and double the height will beA. `10km//s`B. `7km//s`C. `2.50m//s`D. `3.57km//s` |
Answer» Correct Answer - b `(V_(C_(2)))/(V_(C_(1)))=sqrt((M_(2))/(M_(1))xx(r_(1))/(r_(2)))=sqrt((M_(2))/(M_(1))xx((R_(1)+h_(1)))/((R_(2)+h_(2))))` `=sqrt((4xx(R_(1)+h_(1)))/((2R_(1)+2h_(1))))=sqrt((4(R_(1)+h_(1)))/(2(R_(1)+h_(1))))` `V_(C_(2))=sqrt2xx5` `=1.141xx5=7.070"km/s."` |
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