the gravitational force, F, on a rocket at a distance, from the center of the earth is given by F = k / (r^2), where k = 10 ^3 newton*km^2. when the rocket is 10^4 km from the center of the earth, its moving away at .2 km/sec. how fast is the gravitational force changing at that moment? therefore what is the rate of change of the gravitational force in newton/ second?

Question

amistre64 · Answer

rate of change tells me this thing wants derivatives of associated equations

amistre64 · Answer

the gravitational force, F, 
   on a rocket at a distance, r 
         from the center of the earth is given by:

   F = k / (r^2), where k = 10 ^3 newton*km^2. 

when the rocket is 10^4 km from the center of the earth, 
       its moving away at .2 km/sec. 

how fast is the gravitational force changing at that moment? 

therefore what is the rate of change of the gravitational force in newton/ second?

hmm, might be good to draw a picture to make sure we got all the parts

amistre64 · Answer

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amistre64 · Answer

all I can tell is that we should take the derivative of F

F'(r) = (-2)10^3 r^-3 r'

amistre64 · Answer

r' at that moment is .2 soo

i would take a gander and say:

F'(10^3) = -2(10^4)^-3 (10^3) (.2)

$$F'(10^3)=-2(.2)\frac{10^3}{10^{12}}=-2(.2)(10)^{-9}$$

amistre64 · Answer

F =  -4.0 × 10^-10  with any luck; but I cant be too certain that I even got the question right to begin with

anonymous · Answer

you were right with what the question was asking!

amistre64 · Answer

yay!! one of my personalities must be a genuis :)

anonymous · Answer

haha yes you are my savior! but wouldn't you plug in .2 as f'x

amistre64 · Answer

.2 is the rate of change of the radius; r'

anonymous · Answer

because the answer is apparently wrong :(

amistre64 · Answer

by deriving f(r) we get f'(r)* r'

anonymous · Answer

i get that F'(r) = (-2)10^3 r^-3 r', but where did the -2(10^4)^-3 (10^3) (.2) come from?

amistre64 · Answer

i might have transposed some things; let me see

amistre64 · Answer

$$F(r)=kr(t)^{-2}$$
$$F'(r)=-2kr(t)^{-3}*r'(t)$$
$$F'(10^4)=-2k(10^4)^{-3}*.2$$
$$F'(10^4)=-2(10^3)(10^4)^{-3}*2(10^{-1})$$
$$F'(10^4)=-4x(10^3)(10^4)^{-3}(10^{-1})$$
$$F'(10^4)=-4x(10^2)(10^{-12})$$
$$F'(10^4)=-4\ x\ 10^{-10}$$
does that ring more true?

anonymous · Answer

OMG i am sooooo sorry i it's 10 ^ 13 on the question not 10 ^3...  you helped me so much but i told you the wrong thing :((((((((

amistre64 · Answer

:)  then with any luck the concept is sound, just recalculate with the correct value and see if we are good

anonymous · Answer

im so sorry and thank you for helping me i was so stuck! and yes i followed your steps and it was the correct answer!! could you be so kind to help me with my other questions? i really have no luck with optimization problems

amistre64 · Answer

is that the spreading oil tanker one?

anonymous · Answer

yes the gasoline one: Gasoline is pouring into a cylindrical tank of radius 4 feet. When the depth of the gasoline is 6 feet, the depth is increasing at .4 ft/sec. How fast is the volume of gasoline changing at that instant? Round your answer to three decimal places.