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How much liquid hydrogen ( energy density 10^6/litre) is at least needed to bring a satellite of 1 kg in an orbit of 400km. 
Asked by abhipsarout1999 | 14 May, 2020, 15:17: PM
answered-by-expert Expert Answer
I n i t i a l space e n e r g y space o f space t h e space s u r f a c e space o f space t h e space e a r t h T E equals K E plus P E T E equals 0 minus fraction numerator G M m over denominator R end fraction E n e r g y space i n space t h e space o r b i t T E equals K E plus P E T E equals 1 half m v squared minus fraction numerator G M m over denominator R plus h end fraction v equals square root of fraction numerator G M over denominator R plus h end fraction end root T E equals 1 half m fraction numerator G M over denominator R plus h end fraction minus fraction numerator G M m over denominator R plus h end fraction T E equals negative 1 half fraction numerator G M m over denominator R plus h end fraction triangle T E equals negative 1 half fraction numerator G M m over denominator R plus h end fraction minus left parenthesis negative fraction numerator G M m over denominator R end fraction right parenthesis triangle T E equals fraction numerator G M m over denominator R end fraction minus 1 half fraction numerator G M m over denominator R plus h end fraction triangle T E equals G M m open parentheses 1 over R minus 1 half fraction numerator 1 over denominator open parentheses R plus h close parentheses end fraction close parentheses O n space S u b s t i t u t i n g space t h e space v a u e s space w e space g e t T o t a l space e n e r g y equals 3.3 cross times 10 to the power of 7 J e n e r g y space d e n s i t y space i s space g i v e n space b y 10 to the power of 6 L divided by l i t T o t a l space m a s s equals fraction numerator 3.3 cross times 10 to the power of 7 J over denominator 10 to the power of 6 end fraction T o t a l space m a s s equals 33 l t r
Answered by Utkarsh Lokhande | 14 May, 2020, 20:36: PM

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