Asked by Jake
ok I believe this is now correct
0 = (m_M r^2)/m_E - r^2 + 2(3.84403 E 8 m)r - (3.84403 E8 m)^2
how do I solve this algebraically where r^2 is the only unkown
the physics problem
h t t p : / / w w w . j i s k h a . c o m / d i s p l a y . c g i ? i d = 1 2 6 1 6 1 9 5 9 2
0 = (m_M r^2)/m_E - r^2 + 2(3.84403 E 8 m)r - (3.84403 E8 m)^2
how do I solve this algebraically where r^2 is the only unkown
the physics problem
h t t p : / / w w w . j i s k h a . c o m / d i s p l a y . c g i ? i d = 1 2 6 1 6 1 9 5 9 2
Answers
Answered by
MathMate
Things would be less messy if we do the algebraic manipulations before putting in numbers.
Let:
Mass Moon, M2 = 7.35 E 22 kg
Mass Earth, M1 = 5.98 E 24 Kg
Distance Earth to Moon, D = 384,403,000
and r the distance of space craft from the <i>centre</i> of the earth/moon.
GM1m/r² = GM2m/(D-r)²
Cancel out G and m which are identical on both sides, and cross-multiply:
M1(D-r)²=M2r²
Take square-roots on both sides:
√(M1) * (D-r) = √(M2) * r
Solve for r by collecting r terms on the left-hand side:
r(√(M1)+√(M2))=D*√(M1)
or
r=D*√(M1)/(√(M1)+√(M2))
r should be about 9/10 of D.
Let:
Mass Moon, M2 = 7.35 E 22 kg
Mass Earth, M1 = 5.98 E 24 Kg
Distance Earth to Moon, D = 384,403,000
and r the distance of space craft from the <i>centre</i> of the earth/moon.
GM1m/r² = GM2m/(D-r)²
Cancel out G and m which are identical on both sides, and cross-multiply:
M1(D-r)²=M2r²
Take square-roots on both sides:
√(M1) * (D-r) = √(M2) * r
Solve for r by collecting r terms on the left-hand side:
r(√(M1)+√(M2))=D*√(M1)
or
r=D*√(M1)/(√(M1)+√(M2))
r should be about 9/10 of D.
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