A planet with mass m=2.00×1024m=2.00×1024 kg and a second with mass M=5.00×1024

Question

A planet with mass m=2.00×1024m=2.00×1024 kg and a second with mass M=5.00×1024 M=5.00×1024 kg are separated by a distance d=3.51e11 m.

(a) What is the strength of the gravitational force which mm exerts on MM? ____ N ( ± 2E13 N)

(b) What is the strength of the gravitational force which MM exerts on mm? _____ N ( ± 2E13 N)

(c) A third planet with mass m(3)=5.00×1023 kg happens to be midway between MM and mm. What is the net gravitational force (magnitude) on m(3) due to the other two planets?

_____N ( ± 2E13N)

(d) Is the net force toward or away from the smaller mass (mm)?

(e) Where could the third planet be positioned (distance from the larger planet MM) so that the net gravitational force is zero? ____ m ( ± 2E9 m)

Explanation / Answer

A) Force on M is F1 = G*m*M/r^2 = (6.67*10^-11*2*10^24*5*10^24)/(3.51*10^11)^2 = 5.41*10^15 N

B) Force on m is F2 = 5.41*10^15 N

C) Force on m(3) due to m is F13 = (6.67*10^-11*2*10^24*5*10^23)/(3.51*10^11/2)^2 = 2.16*10^15 N

Force on m(3) due M is F23 = (6.67*10^-11*5*10^24*5*10^23)/(3.51*10^11/2)^2 = 5.41*10^15 N

net force is Fnet = (5.41-2.16)*10^15 = 3.25*10^15 N

d) away from the smaller mass

e) F13' = F23'

(6.67*10^-11*2*10^24*5*10^23)/(3.51*10^11-x)^2 = (6.67*10^-11*5*10^24*5*10^23)/(x)^2

2/(3.51*10^11-x)^2 = 5/x^2

sqrt(2)/(3.51*10^11-x) = sqrt(5)/x

1.414*x = 2.23*3.51*10^11-2.23x

x = 2.14*10^11 m from the M

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