I Really Don't Understand These Gravitational Force Questions

[Bees_go_lizzz]

The gravitational force between two objects (mass1 = 14 kg, mass2 = 7 kg) is measured when the objects are 20 centimeters apart. If the distance between them is increased to 40 centimeters, how does the new gravitational attraction compare to the first one that was measured?

The gravitational force between two objects (mass1 = 8 kg, mass2 = 4 kg) is measured when the objects are 10 centimeters apart. The 8 kg mass is replaced with a 4 kg mass and the 4 kg mass is replaced with a 8 kg mass. They are then moved 20 centimeters apart. How does the new gravitational attraction compare to the first one that was measured?

The gravitational force between two objects (mass1 = 8 kg, mass2 = 4 kg) is measured when the objects are 40 centimeters apart. The distance between them is decreased to 10 centimeters, and the 8 kg mass is replaced with a 16 kg mass. How does the new gravitational attraction compare to the first one that was measured?

The gravitational force between two objects (mass1 = 8 kg, mass2 = 4 kg) is measured when the objects are 20 centimeters apart. If the 8 kg mass is replaced with a 32 kg mass and the 4 kg mass is replaced with a 12 kg mass. Additionally, the distance between them is increased to 40 centimeters, how does the new gravitational attraction compare to the first one that was measured?

Thank you so much! Can you please show working so that I don't get confused pleased?

[bhg481]

F = G x (m1 x m2) / d**2

case1
F1 = G x 14 x 7 /20**2 = G x98/400 then
F2 = G x 14 x 7 / 40**2 = G x 98/ 1600 = F1/4

ie F2 = F1 /4 ie, if the separation is doubled, the attactive force is 1/4 of the original.

Use the same formula to calculate the other two cases and tell me your conclusion. If you are wrong I will correct you.
If I do it for you you will learn nothing.

[bhg481]

Incidentally, I am aware that I have used cm rather than metres for the separation - it doesn't affect the principle or the answer, but your supervisor may query it with you.

[Bees_go_lizzz]

Thank you for answering. I did the third one. I got 2.5 kg is smaller is that correct?

[bhg481]

No, the distance between the objects is 1/4 of the original (40 / 10), so the force is 16 x bigger.
Also the 8Kg is replaced by a 16Kg mass = twice as big.

Therefore the attractive force is 16 x 2 = 32 times bigger.

[Bees_go_lizzz]

I am really lost with the last one. I think I went wrong somewhere. I got 174 kg? I really don't think that is correct .

[bhg481]

You start off with 8 kg and 4kg and end up with 32kg and 12kg, so the attraction due to the mass change is (8 x 4) compared to (32 x 12) ie, 12 times greater.
The separation is increased from 20cm to 40cm ie, doubled, so the attraction is only 1/4 of the original.

So, the net change is 12/4 = 3 times the original.

using the formula: F = Gx m1 xm2 /d**2
F1 = G x 8 x 4/20**2 = G x 32/400.

F2 = G 32 x 12 / 40**2 = G x 384/1600

F1/F2 = 1/3

Note that the questions are asking for change (eg, double, half etc) , so the actual values do not matter. The fact that I am using cm for separation, rather than metres (for the MKS system) does not affect the answer to the question.

[Bees_go_lizzz]

Thank you so much!

[bhg481]

The main thing is that you understand what is going on.
The gravitational attraction is proportional to the product of the masses, so if you double one mass you get twice the attraction, if you double one and treble the other you get 2 x 3 = 6 times the attraction.
The force is also INVERSELY proportional to the square of the separation. So, if you double the separation you get 1/4 of the force. If you triple it you get 1/9 of the force etc.
Perhaps it's worth looking at your questions again in that light. eg, Q1 the masses are unchanged, the separation has been doubled, so the attraction has been quartered.

[Peter Pedant]

The first one is easy

as with so much gravitation is inverse sqaure so if you have doubled the distance you have quartered the attraction

I noticed cm and I think you are required to convert to m
and then the attraction will be in N innit ?

I mean you are repeatedly using F = G. m1 m2 /r2

each time with nothing else....

[Peter Pedant]

You have said "I get 174 kg"

NO - kg is a unit of mass

Force in Si units ( thank god ) go around in Newtons

( a force of one newton accelerates one kg at one m/s2 - newtons 2 law )

so the attraction formula F = G . m1m2/r2 should churn out an answer in N

I think your answer should be in the form - the original attraction is x N and then it goes up/down to y N

This question is not a big deal - I last did it in 1968 !