Inverse Power Law: Solve for Power at 400m

In summary, the homework problem asks for the power at 400m if the power at 100m is 5W, but the student is confused because the two concepts are being mixed up. The student explains that they are calculating power loss in a city environment, and that inverse square law doesn't work in practice so they are using 3.5 power law.
  • #1
mrdman
3
1
Homework Statement
If the power at 100m is 5W, using inverse 2.5 power law, find the power at 400m.
Relevant Equations
1 / d ^ x
Hi everyone! Awesome forum!
I'm doubting myself on a problem about inverse square law.
I'll change the actual values from my homework problem as I want to check that I have the right idea rather than the specific numeric answer.

If I am using an inverse 2.5 power law and know the power at 100m is 5W.. If I want to find out the power at 400m, is the following correct...?

100m: 5W
200m: 5 * 1/2 ^ 2.5
300m: 5 * 1/3 ^ 2.5
400m: 5 * 1/4 ^ 2.5

So the answer will be: 5 * (0.25)^2.5
Which is 5 * 0.03125 = 0.15625W

I've had three attempts so far and keep getting varying answers each time!
 
Physics news on Phys.org
  • #2
Looks right to me.
 
  • Like
Likes mrdman
  • #3
Awesome! Many thanks.
 
  • #4
mrdman said:
Homework Statement: If the power at 100m is 5W, using inverse 2.5 power law, find the power at 400m.
Homework Equations: 1 / d ^ x

If I am using an inverse 2.5 power law
Sorry for my dumb question, but this seems a bit confusing to me. There are two kinds of "power" that seem to be getting mixed in this question.

There is the decrease in radiant power from a source, which varies as the inverse of the distance squared (not to the power 2.5).

And there is the "power law" which refers to when a quantity varies with a constant exponent relationship:

https://en.wikipedia.org/wiki/Power_law
1566568375054.png


The constant exponent does not have to be "2" in general, and varies depending on the phenomena being described by the equation.

To me, it is very confusing to mix both concepts together in this question. It implies that somehow radiant power can drop off by the exponent 2.5 instead of 2. In the far field at least, I don't think that can happen. Can it?
 
  • Like
Likes Delta2
  • #5
Hi.. I'm not 100% sure but it may help if I explained the context in which I am doing this calculation.
It is regarding power loss from a cell tower in a city environment. In this environment inverse square law doesn't work out in practice (city obstacles etc) so we were told to use 3.5 power law (I may be getting my terms mixed up. Is it called 3.5 power law?) Above in my example I changed it to 2.5 and changed all the values so I didn't get help with the specific answer, but rather the concept of how to solve it.

Hope that helps and I've not mixed up terms too much.
 
  • Like
Likes berkeman

Related to Inverse Power Law: Solve for Power at 400m

1. What is the Inverse Power Law?

The Inverse Power Law is a mathematical relationship that describes the relationship between two variables, where one variable is inversely proportional to the other. In other words, as one variable increases, the other decreases at a predictable rate.

2. How is the Inverse Power Law used in science?

The Inverse Power Law is used in many fields of science, such as physics, biology, and engineering. It is particularly useful in studying systems that involve forces, such as gravity and electricity. It can also be used to model phenomena such as population growth and the spread of diseases.

3. How is the Inverse Power Law solved for power at 400m?

In order to solve for power at 400m using the Inverse Power Law, you will need to know the values of the two variables involved in the relationship. These variables could be distance and force, for example. Once you have these values, you can use the equation P = k/d^n, where P represents power, k is a constant, d is distance, and n is the power law exponent. Plug in the values and solve for P to find the power at 400m.

4. What is the significance of solving for power at 400m?

Solving for power at 400m can provide valuable information about the relationship between the two variables involved. It can help us understand the behavior of systems and make predictions about how they will change over time. This information can be used in various fields, such as designing structures, improving energy efficiency, and studying natural phenomena.

5. Are there any limitations to the Inverse Power Law?

While the Inverse Power Law is a useful tool in many scientific fields, it does have some limitations. It assumes a linear relationship between the two variables, which may not always be the case. It also requires accurate and precise measurements of the variables involved. Additionally, the power law exponent may vary in different situations, making it necessary to determine the appropriate value for each scenario.

Similar threads

  • Mechanics
Replies
5
Views
1K
  • Electrical Engineering
2
Replies
38
Views
1K
  • Classical Physics
3
Replies
103
Views
3K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Precalculus Mathematics Homework Help
Replies
13
Views
2K
  • Astronomy and Astrophysics
Replies
3
Views
2K
  • Precalculus Mathematics Homework Help
Replies
6
Views
887
Replies
18
Views
7K
  • Introductory Physics Homework Help
Replies
2
Views
820
Back
Top