Radical equations: solve w=Cr^-2 for r; find C if w=100 at

[kristier26]

Many people know that the weight of an object varies on different planets, but did you know that the weight of an object on earth also varies according to the elevation? In particular, the weight of an object follows this equation- w= Cr^-2, where C is a constant, and r is the distance that the object is from the center of the earth.

a. Solve the equation w=Cr^-2 for r

B. Suppose that an object is 100 pounds when it is at sea level. Find the value of C that makes the equation true. ( Sea level is 3,963 miles from the center of the earth.)

So far For a. I have figured to solve for r I must divide by C to get r by itself? this would be w=Cr^-2 = w/C = Cr^-2/C. Is this how I do this?

Then on b. I think I need to substitute 100 for w, but im not sure where to go from there

 

[Deleted member 4993]

Re: Radical equations

Many people know that the weight of an object varies on different planets, but did you know that the weight of an object on earth also varies according to the elevation? In particular, the weight of an object follows this equation- w= Cr^-2, where C is a constant, and r is the distance that the object is from the center of the earth.

a. Solve the equation w=Cr^-2 for r

B. Suppose that an object is 100 pounds when it is at sea level. Find the value of C that makes the equation true. ( Sea level is 3,963 miles from the center of the earth.)

So far For a. I have figured to solve for r I must divide by C to get r by itself? this would be w=Cr^-2 = w/C = Cr^-2/C. Is this how I do this?

yes and No...

You need to have it following form - r on the left hand-side

r = ???

\(\displaystyle w \, = \, \frac{C}{r^2}\)

\(\displaystyle r^2 \, = \, \frac{C}{w}\)

Now continue...
Then on b. I think I need to substitute 100 for w, but im not sure where to go from there

and r = 3963

then evaluate 'C'

 

[kristier26]

would I use the original equation of w=Cr^-2 or the other from part a. r^2=c/w to plug in the numbers from part b?

 

[Deleted member 4993]

would I use the original equation of w=Cr^-2 or the other from part a. r^2=c/w to plug in the numbers from part b?

It does not matter which equation you use - those are equivalent.

Just remember that you are trying to solve for 'C'.

 

[mmm4444bot]

It does not matter because both equations are equivalent. In other words, they both state the same relationship between C, r, and w.

You could also first solve either one of them for C, and then substitute your values for r and w into the result.

All three equations will give the same value for C.