if the mass of earth is greater than apple then what happens if another planet with gravitational attraction is close to earth
Answers
Earth’s mass (rounded) is 6 x 10^24 kg. Its radius is approximately 6,378,100 m. In the above equations, adding the extra 200 meters between objects and computing, we get:
Values for ball 1 (M1a) (the plastic bowling ball) vs Earth
Fg = G*M1a*M2/R^2
Fg = [6.67 x 10^-11] * 5 kg * 6 x 10^24 kg /6,378,300 m^2 = 49.18551287 force units
Acceleration of M1a equals A = F/M1a = 49.18551287 / 5 kg =
9.837102574 m/sec^2
Values for ball 2 (M1b) (the bigger uranium bowling ball)
Fg = G*M1b*M2/R^2
Fg = [6.67 x 10^-11] * 400 kg * [6 x 10^24] kg /6,378,300 m^2 = 3934.841029 force units
Acceleration of M1b equals A = F / M1b = 3934.841029 / 400 kg =
9.837102574 m/sec^2
Values for combined mass (M1c) (both balls) Fg = G*M1c*M2/R^2
Fg = [6.67 x 10^-11] * 405 kg * [6 x 10^24] kg /6,378,300 m^2 = 3984.026542 force units
Acceleration of M1b equals A = F / M1c = 3984.02642 / 405 kg =
9.837102574 m/sec^2
Values for mass 2 (M2) (Earth) moving toward the combined masses M1a and M1b
Fg = G*M1c*M2/R^2
Fg = [6.67 x 10^-11] * 405 kg * [6 x 10^24] kg /6,378,102 m^2 = 3984.026542 force units
Acceleration of M2 equals A = F / M2 = 3984.026542 /6x10^24 =
6.0640 x 10^-22 meters/sec^2