Physics (Gravitational Force) Homework Question?

Question

I've got a physics test coming up, and right now I'm studying for it using my homework, but I'm confused with two of these questions regarding F = G*m1*m2/R2. Can anyone please answer them and explain them?

1. Two objects of masses around m1 and m2 are initially separated at a distance of 5 cm. The distance is then changed to 10 cm. What is the ratio of gravitational force experienced by both objects at the initial condition and that at the final condition?

2. Three equal bodies are found at the corners of an equilateral triangle. If the force between two of three bodies if F, the magnitude of force experienced by each body is?

Seamus O · Accepted Answer

1. G m₁ m₂ is the same for both distances ... so it doesn't change

when the distance between the masses is doubled the denominator changes from r² to (2r)² = 4r²

so the gravitational force at the greater distance is 1/4 as strong

so the gravitational force when the distance between the masses is 5 cm is 4 times as strong as the gravitational force when the distance is increased to 10 cm

the ratio they want is (gravitational force at 5 cm separation) / (gravitational force at 10 cm separation) = 4 / 1 = 4 : 1




2. The triangle is equilateral ... so all sides are of equal length and all angles equal 60°

Call the three bodies A, B and C

The resultant gravitational force on A due to B and C is the vector sum of the two separate gravitational forces


in general the easiest way to add vectors is to do it in component form so I'll do it that way to show the method:

imagine little x-y axes centered on A

The gravitational force pointing from A to C makes an angle of 180 + 60 = 240° with the +x axis

so the gravitational force of C ON A = F cos 240 i + F sin 240 j ... [where i is the horizontal unit vector and j is the vertical unit vector]


the gravitational pointing from A to B makes an angle of 360 - 60 = 300° with the +x axis

so the gravitational force of B ON A = F cos 300 i + F sin 300 j



Adding the two separate vectors:

resultant gravitational force on A = (F cos 240 + F cos 300)i + (F sin 240 + F sin 300)j

= [(-1/2)F + (1/2)F]i + [(-√3 / 2)F + (-√3 / 2) F]j

= 0i - √3 F

so the magnitude of the force experienced by A = √3 F

b/c of the symmetry of the triangle ... the force experienced by B and C has the same magnitude 

so the magnitude of the force experienced by each body is √3 F





btw ... if you are interested there is a quicker way to get the required force for this question ... the resultant force on A will be directed away from A and bisect the line from B to C midway between B and C ... using geometry you get a vector triangle with 2 sides of length F and the angle between them of 120° ... then using the law of cosines you can find the magnitude of the other side and that is the magnitude of the resultant force on A ... and b/c of symmetry for B and C as well ... but I prefer the component method for most vector stuff because it is usually quicker ... especially when you have to add more than two vectors and get both the magnitude and the direction

? · Answer

Hello Tim, given mass m = forty two kg Also Mm / Me = a million/eighty one.three and Rm / Re = zero.27 Now drive on the planet is bought by means of the components m*ge ge = nine.eight m/s^two So drive at the mass on the planet = 411.6 N Now remember the expression g = GM/R^two So gm / ge = Mm/Me * (Re/Rm)^two Hence you'll be able to uncover gm = ge * a million/eighty one.three * (a million/zero.27)^two Simplifying we get gm = a million.sixty five m/s^two (just about) Hence the drive at the identical mass by means of the moon is bought by means of m*gm = forty two*a million.sixty five = sixty nine.three N As g importance at the moon is nine.eight/a million.sixty five ie five.ninety four instances smaller than that on the planet, the drive at the mass too can be just about 6 instances smaller.

Physics (Gravitational Force) Homework Question?

2 Answers