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

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SUMMARY

The gravitational force exerted by the Moon on a 50-kg student is calculated using the formula F = G * m1 * m2 / d², where G is the gravitational constant (6.67 x 10^-11 N m²/kg²), m1 is the mass of the student (50 kg), m2 is the mass of the Moon (7.35 x 10^22 kg), and d is the distance from the Moon to the student (3.84 x 10^8 m). The correct calculation yields a gravitational force of approximately 0.0016623433 Newtons. Both the student and the Moon attract each other with equal force, confirming Newton's third law of motion.

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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
 
Last edited:
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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?
 
Last edited:
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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