Question

ATT Myininaya:
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 = 10^13 newton • km2. When the rocket is 10^4 km from the center of the earth, it is moving away at 0.2 km/sec. How fast is the gravitational force changing at that moment? Give units. (A newton is a unit of force.)

Answer 1

so it wants to know how fast F is changing
so we need to find F'

Answer 2

is this calculus or physics?

Answer 3

Calculus

Answer 4

it looks like k is a constant

Answer 5

oh maybe

Answer 6

that is where I may have gone wrong

Answer 7

This is a related rates problem

Answer 8

you may have to do df/dt

Answer 9

\[F'=\frac{0 \cdot r^2-k \cdot 2rr'}{r^4}=\frac{-2kr'}{r^3}\]

Answer 10

so we are given everything

Answer 11

huh?

Answer 12

i found F'

Answer 13

I got a diff answer

Answer 14

F ' =-2k/r^3

Answer 15

i used the product rule

Answer 16

\[(r^2)'=2r^{2-1}r'=2r^1r'=2rr'\]

Answer 17

we can use the product rule if you want

Answer 18

I just used the quotient rule and still got a diff answer

Answer 19

\[F=kr^{-2} > F'=(-2)kr^{-2-1}r'=-2kr^{-3}r'=\frac{-2kr'}{r^3}\]

Answer 20

what is r'?

Answer 21

r is a function of time

Answer 22

I am a drop confused

Answer 23

r' is the rate that the distance is changing from the center of the earth

Answer 24

\[[r(t)]'=r'(t)\]

Answer 25

oh it is dr/dt?

Answer 26

We didn't derive it normally we are doing like the related rate format

Answer 27

these mean the same thing
\[r'(t)=\frac{dr}{dt}\]

Answer 28

just different notation

Answer 29

k thanks

Answer 30

ya i was just getting confused a lil

Answer 31

one is i think newton's notation
and the other is lebniz notation i believe

Answer 32

but they mean the samething

Answer 33

\[F'(t)=[k \cdot [r(t)]^{-2}]'=k[[r(t)]^{-2}]'=k(-2)[r(t)]^{-2-1}r'(t)=-2kr'(t)[r(t)]^{-3}\]

Answer 34

oh. i get it. I guess my book only shows liebniz's notation

Answer 35

maybe
i usually use newton's notation
but that doesn't mean lebniz's is inferior notation

Answer 36

ya i like newton's notation better but my book shows me this way df/dx ...

Answer 37

is df/dt=-2(10^9)*.2?

Answer 38

\[r=10^4 km ; r'=.2 km/\sec ; k=10^{13} newton \cdot km2\]

Answer 39

oh whoops i made a mistake

Answer 40

i forgot the cubed in the denomanator

Answer 41

df/dt=2(10)*.2

Answer 42

which is 8?

Answer 43

\[F'=\frac{-2 k r'}{r^3}=\frac{-2(10^{13} N/km^2) \cdot (0.2 km/\sec)}{(10^4 km)^3}\]
\[=\frac{-2 (10^{13}N/km^2) \cdot (\frac{2}{10} km /\sec)}{10^{4 \cdot 3} km^3}\]
\[=\frac{-4 (10^{12} N/(km \cdot \sec))}{10^{12 }km^3}\]
\[=-4 \frac{N}{km \cdot \sec} \cdot \frac{1}{km^3}=-4 \frac{N}{km^4 \sec}\]
what did i miss?

Answer 44

i think I went wrong somewhere but I will have to recheck it

Answer 45

and maybe r' should have been negative since is says moving away

Answer 46

and therefore F'= +4 not -4

Answer 47

I may have punched accidentally the wrong number into the calculator and that is y i got the wrong number

Answer 48

I think it is positive

Answer 49

yes i think so too

Answer 50

since the further away it is the larger r is

Answer 51

oh wait are you talking about F'
because thats what i mean

Answer 52

r' is negative
F' is positive

Answer 53

yes that is what I meant

Answer 54

meaning it shldn't be -.2

Answer 55

Thanks for ur help

Answer 56

Most ppl on here can't answer these questions. There are only a few ppl and hey r not on at all times

Answer 57

now that i have read it couple of times
r' is increasing not decreasing so r' is positive
so F' is negative

Answer 58

y is that?

Answer 59

meaning that r'=-.2?

Answer 60

no increasing would mean r'=.2

Answer 61

decreasing means r'=-.2

Answer 62

right so that is what we have right now

Answer 63

but the distance is getting greater so r'=positive .2

Answer 64

ok

Answer 65

yes thats how i solved the problem initially

Answer 66

Thanks;-D

Answer 67

F'=-4
this means the gravitationial force is decreasing at a rate of 4 N/km^4 per sec

Answer 68

if you wanted to know how to interpret that

Answer 69

I know u r gonna kill me but isn't it km^3

Answer 70

oh i see it

Answer 71

\[F'=\frac{-2 k r'}{r^3}=\frac{-2(10^{13} N/km^2) \cdot (0.2 km/\sec)}{(10^4 km)^3} \]
if i were to mess with the units i would have
\[\frac{\frac{N}{km^2} \cdot \frac{km}{\sec}}{km^3}=\frac{\frac{N}{km \cdot \sec}}{km^3}\]
\[=\frac{N}{km \cdot \sec} \cdot \frac{1}{km^3}=\frac{N}{km^4 \cdot \sec}\]

Answer 72

OMG i wld not be able to solve that on my own!!!! I wld have totally left that out. I don't know how to solve wordproblems properly

Answer 73

i usually don't bother with the units

Answer 74

but they are just like numbers
all you have to is treat them like numbers i mean

Answer 75

well we had to write it in u

Answer 76

units. Ok thanks for ur help. I really appreciate it. What level math r u in?

Answer 77

oh wow good luck

Answer 78

i need to go back and read some stuff

Answer 79

i have alot of work to do lol

Answer 80

y didn't u go into math u really good?

Answer 81

i got my masters in mathematics

Answer 82

i think thats enough

Answer 83

ya I am aiming for that to. I am just in the beggining

Answer 84

you will get there and you will probably be tons smarter than me by then

Answer 85

If i ever need help will u be able to help me?

Answer 86

if i'm around, i would be happy to help

Answer 87

Oh my u sound like my dad already!!!

Answer 88

Nice to meet you

Answer 89

i hope i will never be anyone's dad
since i'm a chick lol