What is the Gravitational Force Between the Moon and a 50-kg Student?

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The discussion revolves around calculating the gravitational force exerted by the Moon on a 50-kg student using the formula F = G * m1 * m2 / d^2. The initial calculation yielded an incorrect result, prompting participants to reassess the values used, particularly the squaring of the distance. After correcting the calculation, the gravitational force was determined to be approximately 0.0016623433 Newtons. It was clarified that both the Moon and the student attract each other with equal force, consistent with Newton's third law of motion. The conversation emphasizes the importance of accurate calculations in physics problems.
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Homework Statement



It is said that people often behave in unusual ways during a full moon. a) Calculate the gravitational force that the moon would exert on a 50.0-kg student in your physic's class. The moon is 3.84 x 108m from the Earth and has a mass of 7.35 x 1022kg. b) Does the moon attract the student with a force that is greater than, less than, or the same as, the force with which the student attracts the moon?

Homework Equations



F = G times m1 times m2 divided by d squared

or

F = G * m1 *m2 / d^2

The Attempt at a Solution

A) The force of attraction or the gravitational force between two objects is the product of the two masses divided by the distance between the two objects squared, all multiplied by the gravitational constant.

F = the force of attraction between two masses
G = gravitational constant
m1, m2 = the two masses
d = the distance between the two masses

F = G times m1 times m2 divided by d squared

or

F = G * m1 *m2 / d^2

so F = (6.67*10^-11)*(50)*(7.35*10^22) / (3.84*10^8)^2 then

F = 0.0016623433

B) They both attract each other the same amount
 
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The answer to (a) is incorrect. Try plugging in the numbers again.
 
kuruman said:
The answer to (a) is incorrect. Try plugging in the numbers again.

Is A correct now? when i originally posted i put in 9. something by accident is the 6.38 correct?
 
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No, 6.38 (I assume Newtons) is not correct. Like I said, redo the calculation. Look at your expression, the numerator is on the order of 1014 and the denominator is on the order of 1016. You can't get an answer n the order of 1.
 
kuruman said:
No, 6.38 (I assume Newtons) is not correct. Like I said, redo the calculation. Look at your expression, the numerator is on the order of 1014 and the denominator is on the order of 1016. You can't get an answer n the order of 1.

Thanks for the help, I realize that when squaring the denominator i was squaring just the ^8 instead of the whole problem and now that i fixed that I got

F = 0.0016623433
 
That's about right.
 
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