# Tough assignment question - help would be appreciated

• Mohammed17
In summary, the ratio (r)max/(r)min, where (r)max is the moon's maximum distance from the center of the planet and (r)min is the minimum, is 0.95. This can be found by dividing (r)max by (r)min and then taking the square root of the result. This is because the gravitational force exerted on the moon by the planet is 11% larger at its maximum distance compared to its minimum distance, and this is inversely proportional to the distance.
Mohammed17
As a moon follows its orbit around a planet, the maximum gravitational force exerted on the moon by the planet exceeds the minimum gravitational force by 11%. Find the ratio (r)max/(r)min where (r)max is the moon's maximum distance from the center of the planet and (r)min is the minimum

Ok so I know the equation that will be used is:

Fg = (G)*m1*m2 / r^2

But I have nooooo idea where to start.

Do I do Fg of the moon = (0.11)(Fg)planet + (Fg)planet

...?

You're on the right track, but you need to distinguish between Fg((r)max) and Fg((r)min) in the equation that you wrote.

Ok so it would be:

1/r^2 (max) = (0.11)/r^2 + 1/r^2

so it would be: 1/r^2 max = 1.11/r^2 min

cross multiplication would give me:

(1.11)r^2 max = r^2 min

Since we need the ratio of r^2 max to r^2 min
divide both sides by r^2 min.

1.11 r^2 max / r^2 min = 1

now divide both sides by 1.11
r^2 max / r^2 min = 1/1.11 = 0.90

Square root of (r^2 max / r^2 min) = Square root of (0.90)

Therefore:

(r)max / (r) min = 0.95

Last edited:
Mohammed17 said:
Ok so it would be:

1/r^2 (max) = (0.11)/r^2 + 1/r^2

so it would be: 1/r^2 max = 1.11/r^2 min

cross multiplication would give me:

(1.11)r^2 max = r^2 min

Since we need the ratio of r^2 max to r^2 min
divide both sides by r^2 min.

1.11 r^2 max / r^2 min = 1

now divide both sides by 1.11
r^2 max / r^2 min = 1/1.11 = 0.90

Square root of (r^2 max / r^2 min) = Square root of (0.95)
(r) max / (r) min = 0.95
Is this correct?

May someone go over it. Is my logic correct? I guess since Fg is inversely proportional to radius, then if Fg is 11 % larger, the radius must be a portion smaller ?

Last edited:
Yes, if the gravitational force is larger, the distance is smaller by a factor involving an inverse square root.

fzero said:
Yes, if the gravitational force is larger, the distance is smaller by a factor involving an inverse square root.

is my value correct though? 0.95 ?

What you wrote looks correct but the problem is asking for (r)max/(r)min.

fzero said:
What you wrote looks correct but the problem is asking for (r)max/(r)min.
I did get (r) max / (r) min. IT was equal to 0.95. I just forgot to show that i square rooted both sides in the above posts. I fixed that though.

Last edited:

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