Calculus Problem?

Question

I have this problem for a calculus assignment and i can't get my head around it since it looks like a physics problem.

Newtons Law of gravitation says that the magnitude F of the force exerted by a body of mass m on a body of mass M is given by

F= G*m*M / r^2

where G is the gravitational constant and r the distance between the bodies. A cosmonaut inside a spaceship is approaching a newly discovered Planet Htrea. Looking at his instruments, he notices a distance of 2000 km from the planet, its gravitational pull is increasing at 1 N/Km as he approaches the planet. Compute the gravitational force that the planet will exert on the spaceship at a distance of 10000 KM from it.

Answer 1

Start by differentiating the force equation:f`=-2GmM/r^3; therefore, when r=2000,f`=1=(-2GmM/2000^3)
Therefore, 4000000000=-GMm
and M=4000000000/Gm
Now substitute this back into the original equation:f(10000)=G*m*(4000000000/Gm)/100000000=40N

Answer 2

Is all the information given? If you know the initial gravitational force you can find out what it will be at a new distance of r by solving the following eqaution.


dF/dr= 1 N/m

dF/dr= (F final- F initial)/( R final- R inital)= 1 N/m

where R final- R initial is 10000-2000=8000

Answer 3

G, M, and m are all constants, so you just need to use the given data to work out the value of G*m*M, and then substitute
r = 10000 in the formula to find F.

It's a bit like the previous one you asked. Differentiate to get
F' = -2*G*m*M/r^3.

The minus sign just represents the fact that F is increasing when r is decreasing, so ignore it and substitute
F' = 1 when r = 2000.

Solve to find the value of G*m*M, then put r = 10000 in the original formula to find F.

Answer 4

f (x) = x^3 - 12x + 1 . . . the first derivative set to 0 finds turning or stationary points f ' (x) = 3x^2 - 12 3x^2 - 12 = 0 3 * (x + 2) * (x - 2) = 0 x = 2 ... x = - 2 . . . the second derivative evaluated at x = 2 and -2 determines if those points are min, max, or neither. f ' ' (x) = 6x f ' ' (2) = 6*2 = 12 <== positive value indicates x=2 is a local minimum f ' ' (-2) = 6*(-2) = -12 <== negative value indicates x=-2 is a local maximum a.) x = - 2 is a maximum, and x=2 is a minimum ... so x = - infinity to -2 is increasing x = -2 to +2 is decreasing x = +2 to + infinity is increasing b.) f (-2) = (-2)^3 - 12*(-2) + 1 = 17 f (2) = (2)^3 - 12*(2) + 1 = - 15 c.) . . . the second derivative set to 0 finds inflection points, or where concavity changes 6x = 0 x = 0 <=== inflection point x = - 2 is a maximum, so must be concave down concavity changes at the inflection point(s) ... so x = - infinity to 0 is concave down x = 0 to + infinity is concave up

Answer 5

yes this is a physics formula, but you can still do it! Here ill tell you what the values in the problem corrospond to:

F = 1N
G m M= what ur finding
r = 2000 km

pretend G,m, and M is one quantity

first find G,m, and M for 2000km

then use that number to find 10000km.

Pretty easy stuff

Answer 6

It is a physics question but it is a simple differential problem.

F = G*m1*m2/r^2
G is a constant, m1 is constant, m2 is constant. Only r is changing
R is the radius of the planet. But for this problem I think they assuming R = 0. To truly figure out R you need another data point.
F'@2000km+R = 1 N/Km = -2*G*m1*m2/r^3
F'@10000km+R = 0.008 N/Km.

Answer 7

F=mMG/ r^2
dF/dr=-2mMGr^-3
when r = 2500 km dF/dr = 1
by inspection
(dF/dr) is proprtional to r raised to the negative 3
Therefore
1 *(2500^3) = x *(10000^3)
x= 0.25^3 = 0.015625 N/km

Answer 8

dF/dr=-2G*m*M/r^3=(-2/r)*G*m*M/r^2
plugging in the vaues given
G*m*M/r^2(-2/r)=dF/dr
=>G*m*M=1*2000^3/-2
when r=10000
G*m*M/r^2=-2000^3/2(10000^2)
=-2000*2000*2000/2*10000*10000
=--40