How far from Venus must a space probe be along a line toward the Sun so that the?

How far from Venus must a space probe be along a line toward the Sun so that the Sun's gravitational pull on the probe balances Venus's pull?|||a(V) = a(S) = Gm(V)/d(V)^2 = Gm(S)/d(S)^2


Thus d(S)/d(V) = sqrt(m(S)/m(V))


d(V)+d(S) = d(StoV) (whose value we know)


d(S)/d(V)+1 = d(StoV)/d(V) = sqrt(m(S)/m(V))+1


d(V) = d(StoV)/(sqrt(m(S)/m(V))+1) (answer)


I get d(V) = 1.0821E11/(sqrt(1.9889E30/4.869E24)+1) = 169044805 m


Checking, Gm(V)/d(V)^2 = Gm(S)/d(S)^2 = 0.0113716 m/s^2.


EDIT: Waheed, one reason your result is so huge (way beyond the solar system) is that you need another change in the sign of Y^2, making it .. + Y^2 .. when going from line 6 to line 7. This affects everything that comes after that.


Oops, and the other, major thing. The larger the ms/mv ratio, the larger you want the ds/dv ratio, so your 1st equation, ms / mv = ( Y)^2 / ( 108 200 000 - Y)^2, is inverted.|||Thanks!

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|||Let Y distance from venus;


we have assumed : mass of sun = 1.989 x 10^30 kg


mass of Venus = 4.868 x 10 ^24 kg


distance of venus from Sun = 108 200 000 km





ms / mv = ( Y)^2 / ( 108 200 000 - Y)^2


(1.989 x 10^30 / 4.868 x 10^24) = ( Y)^2 / ( 108 200 000 - Y)^2


0.4085866 x 10^ 6 ( 1.17072 x 10^16 - 216 400 000 Y - Y^2) = Y^2


4.78342 x 10^21 - 8.84181 x 10^13 Y - 408587.6 Y^2 = 0





Y = { 8.84181 x 10^13 + ( 7.81776 x 10^27 + 7.81778 x 10^27 )^0.5}/2


Y = 1.0673 x 10^14 km from Venus ans