Tough assignment question - help would be appreciated

  • Thread starter Mohammed17
  • Start date
  • Tags
    Assignment
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.
  • #1
Mohammed17
10
0
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

...?
 
Physics news on Phys.org
  • #2
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.
 
  • #3
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:
  • #4
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:
  • #5
Yes, if the gravitational force is larger, the distance is smaller by a factor involving an inverse square root.
 
  • #6
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 ?
 
  • #7
What you wrote looks correct but the problem is asking for (r)max/(r)min.
 
  • #8
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:

FAQ: Tough assignment question - help would be appreciated

What does the assignment require?

The assignment most likely requires you to complete a task or solve a problem related to a specific subject, such as a research project or an experiment.

How can I prepare for a tough assignment?

To prepare for a tough assignment, you can start by thoroughly understanding the instructions and requirements. You can also break down the task into smaller, more manageable parts and create a timeline for completing each part. It is also helpful to gather any necessary materials or resources beforehand.

What if I don't understand the assignment?

If you don't understand the assignment, you can reach out to your instructor or classmates for clarification. You can also conduct additional research or seek assistance from a tutor or academic support center.

How do I stay motivated while working on a tough assignment?

To stay motivated, you can remind yourself of the importance of completing the assignment and how it contributes to your overall learning and academic success. You can also take breaks and reward yourself for completing each step of the assignment.

Is it okay to ask for help on a tough assignment?

Yes, it is absolutely okay to ask for help on a tough assignment. Seeking assistance from others can provide valuable insights and help you overcome any challenges or obstacles you may encounter. Just make sure to properly credit any sources or individuals who have helped you with the assignment.

Similar threads

Replies
6
Views
2K
Replies
21
Views
3K
Replies
3
Views
1K
Replies
12
Views
2K
Replies
3
Views
5K
Replies
10
Views
12K
Replies
2
Views
4K
Back
Top