Newton's Law of Gravitation
So I need some help with this problem:
Newton's Law of Gravitation states that the magnitude F of force exerted by a body of mass m on a body of mass M is F(r)=(GmM)/r^2, where G is the gravitational constant and r is the distance between the center of the bodies. Suppose that one of the bodies is the earth and the other is a 1000-kg satellite. The earth's mass is approximately 5.98(10^24) kg concentrated at the center and its radius is approximately 6.37(10^6)m. G=6.67(10^-11) N * m^2/kg^2
a. Determine how fast the force between the earth and satellite changes when the satellite has launched vertically to a height of 1000 km.
b. Explain the physical meaning of the minus sign for the rate of force with respect to the distance between the bodies.
c. The determination of work required, when a force and change in distance is given, is analogous to determining the displacement when the velocity and change in time is given or the change in charge when the current and the change in time is given. Compute the work required to launch the satellite vertically to a height of 1000 km.
I really need some help trying to figure out what I need to do in order to proceed with this problem. Don't want the answers just help solving it
Thanks in advance
Re: Newton's Law of Gravitation
You are given:
=\frac{GmM}{r^2})
a) Differentiate with respect to time
.
b) As
increases, what happens to
.
c) Use \,dr)
Re: Newton's Law of Gravitation
Okay I actually have a question how would I start to differentiate with respect to t?
Re: Newton's Law of Gravitation
would I have to do implicit differentiation with dF/dt?
Re: Newton's Law of Gravitation
I would write the function as:
=GmMr^{-2})
Now, use the power and chain rules (recall
is a constant).
Re: Newton's Law of Gravitation
okay so I got
dF/dt = (-2GmM)/r^3 ? is that right? cause I don't think it is in respect to time t
Re: Newton's Law of Gravitation
You have neglected to apply the chain rule. Your answer would be correct if we are differentiating with respect to r. So, what do you need to add?
Re: Newton's Law of Gravitation
OK I am unsure of what I use the chain rule on.
Re: Newton's Law of Gravitation
Since r is a function of t you want to apply it to r.
Re: Newton's Law of Gravitation
oh got it haha its late lol
So I got:
(-2GmM/r^3)(GmM/r^2)
because t is a function of r so I multiply the derivative by the original function right?
Re: Newton's Law of Gravitation
No, you want to write:
)=-\frac{2GmM}{r^3}\cdot\frac{dr}{dt})
Re: Newton's Law of Gravitation
Oh I actually had that I just thought since it was a function I had to replace it with the original equation
Re: Newton's Law of Gravitation
Are you given any information regarding the rate at which the satellite moves upwards?
Re: Newton's Law of Gravitation
no everything I am given is in my original post
Re: Newton's Law of Gravitation
Okay, then your result for part a) will have to be a function of dr/dt. Just plug in the given distance from the center of the Earth.