On the way from a planet to amoon, astronauts reach a point where that moon’sgra

Question

On the way from a planet to amoon, astronauts reach a point where that moon’sgravitational pull is stronger than that of the planet. The massesof the planet and the moon are, respectively, 5.19E24kg and 7.36E22 kg. The distance from the center of the planet to thecenter of the moon is 3.86E8 m.
Determine the distance of this point from the center of the planet.The value of the universal gravitational constant is6.67259E11 N·m^2/kg^2 Answer in units of m. On the way from a planet to amoon, astronauts reach a point where that moon’sgravitational pull is stronger than that of the planet. The massesof the planet and the moon are, respectively, 5.19E24kg and 7.36E22 kg. The distance from the center of the planet to thecenter of the moon is 3.86E8 m.
Determine the distance of this point from the center of the planet.The value of the universal gravitational constant is6.67259E11 N·m^2/kg^2 Answer in units of m.

Explanation / Answer

Gravitational force from the planet =G.M.m/r2 Where M is mass of the planet and m is mass of the object, andr=distance of the object to the center of the planet From F=ma we can have a1 = GM/r2 (a1= gravitational accelarationdue to planet's gravity) Similarly, gravitational accelaration due to moon'sgravity a2 = GM2/r22 Set a2=a1 >>>Mr22 = M2r2(r2 = D - r, D= distance between 2 centers =3.86e8) Plug in M=5.19e24, M2=7.36e22, andD=3.86e8 and solve the above equation, 5.19e24(3.86e8-r)2 = 7.36e22*r2, takingsquareroot both sides, 2.278e12(3.86e8-r) = 2.713e11*r >>> r = 3.45e8 (m) (distance from that point tocenter of the planet) Set a2=a1 >>>Mr22 = M2r2(r2 = D - r, D= distance between 2 centers =3.86e8) Plug in M=5.19e24, M2=7.36e22, andD=3.86e8 and solve the above equation, 5.19e24(3.86e8-r)2 = 7.36e22*r2, takingsquareroot both sides, 2.278e12(3.86e8-r) = 2.713e11*r >>> r = 3.45e8 (m) (distance from that point tocenter of the planet)

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