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Mathematics 21 Online
OpenStudy (anonymous):

What happens to the gravitational force between two objects if each mass doubles and the distance between then is multiplied by 8

OpenStudy (anonymous):

F = G m1 m2 / r^2 F_2 = G 2*m1 2*m2 / (8r)^2 = 4 G m1 m2 / 64 r^2 = 4/64 * G m1 m2 / r^2 F_2 = 1/16 F The new force will be 1/16 of the old force.

OpenStudy (wolf1728):

Gravity decreases by the inverse square of the distance. Double the distance the gravity decreases by four. Increasing the distance by a factor of 8 reduces the gravity to 1/64 of what it once was. Gravity varies porportionately with mass. You double the mass of each object and the gravity increases by 4 . So the gravity is 4/64 of what it once was. 4/64 = 1/16

OpenStudy (anonymous):

thank you.

OpenStudy (wolf1728):

u r welcome

OpenStudy (anonymous):

2 more questions if your willing to answer

OpenStudy (wolf1728):

okay I'll stick around

OpenStudy (anonymous):

1.)What is the gravitational force between two objects if the mass of one object is doubled, the mass of the second object is tripled, and the distance between them is multiplied by (1/5).

OpenStudy (anonymous):

2.) If the Earth's mass were to double what it is, in what ways would the moon's orbit be different?

OpenStudy (wolf1728):

Gravity decreases by 25 (1/5²) Gravity increases by 2*3 = 6 So, the gravity is 6/25 as weak.

OpenStudy (wolf1728):

Just to make sure I checked that answer with a gravity calculator. http://www.1728.org/gravity.htm It is correct.

OpenStudy (anonymous):

thanks for all the help, much appreciated!

OpenStudy (wolf1728):

Still have to work on the Moon question.

OpenStudy (wolf1728):

Unsure about this Assuming the Earth-Moon distance remains the same: If the Earth's mass is doubled, the Moon would travel 1.4 times faster than it does now. It currently orbits the Earth in about 29.5 days and this time would shorten to about 20.9 days. Only 20.9 days would be required to go from full Moon to full Moon. This is based on a version of Kepler's 3rd Law t² = (4 · π² · r³) / (G • m) To make things easier, let's say everything equals 1 Now if the mass increases to 2 then t² goes from 1 to .5 and t = sqrt(.5) = 0.7071067812 (1/0.7071067812) = 1.4142135623 The tide periods should remain the same because these depend on Earth's rotation (24 hours) and not the Moon's revolution.

OpenStudy (anonymous):

Alright. Thanks for the time and help.

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