PartA.) A planet of mass 5 × 10 24 kg is at location ‹ 3 × 10 11 , 6 × 10 11 , 0

Question

PartA.)

A planet of mass 5 × 1024 kg is at location ‹ 3 × 1011, 6 × 1011, 0 › m. A star of mass 9 × 1030 kg is at location ‹ 5 × 1011, 6 × 1011, 0 › m. What is the force exerted on the planet by the star? (It will probably be helpful to draw a diagram, including the relevant vectors.)

Part B.)

A planet exerts a gravitational force of magnitude 4e22 N on a star. If the planet were 5 times closer to the star (that is, if the distance between the star and the planet were 1/5 what is is now), what would be the magnitude of the force on the star due to the planet?

Explanation / Answer

A. given mass of planet m = 5*10^24 kg
   location r1 = (-3*10^11, 6*10^11,0) m
   mass of star, M = 9*10^30 kg
   location, r2 = (5*10^11, -6*10^11, 0) m
   so a vector pointing towards the star from the planet is r
   r = r2 - r1 = (5*10^11, -6*10^11, 0) - (-3*10^11, 6*10^11,0) = (8*10^11, -12*10^11, 0) m
   magnitude of this vector
   |r| = 10^11*sqroot(8^2 + 12^2) = 14.422*10^11 m
   so a unit vector in this direction is given by
   r' = r/|r| = (0.5547,-0.832,0)
   hence force
   F = GMm*r'/|r|^2 = 6.67*10^-11*9*5*10^54((0.5547,-0.832,0))/(14.422*10^11)^2
   F = (0.8004,-1.2006,0)*10^21 N

B. given F = 4*10^22 N
   when distance is reduced by a factor of 5
   the force will increase to a factor of 25 as the force of gravitation follows an inverse squared law
   hence
   F' = 25*4*10)^22 = 10^24 N

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