I I want to see how d/dt (-mu/r) = mu/r^2 * v

[AdrianGriff]

So here it says that $d/dt [-\mu/r] = (\mu/r^2)* v$ where $r$ is the separation or radius, and $v$ is the velocity, and $\mu$ is a parameter equal to $G(M+m)$

And I do not see how they get this to be equal, because all of my tries using the quotient rule create another term that they do not have. Any help will be appreciated :)

 

[fresh_42]

So here it says that $d/dt [-\mu/r] = (\mu/r^2)* v$ where $r$ is the separation or radius, and $v$ is the velocity, and $\mu$ is a parameter equal to $G(M+m)$

And I do not see how they get this to be equal, because all of my tries using the quotient rule create another term that they do not have. Any help will be appreciated :)

Do you know what $\dfrac{d}{dt} f(t)^{-1}$ is?

 

[AdrianGriff]

Do you know what $\dfrac{d}{dt} f(t)^{-1}$ is?

Haha, no I do not :)

 

[PeroK]

So here it says that $d/dt [-\mu/r] = (\mu/r^2)* v$ where $r$ is the separation or radius, and $v$ is the velocity, and $\mu$ is a parameter equal to $G(M+m)$

And I do not see how they get this to be equal, because all of my tries using the quotient rule create another term that they do not have. Any help will be appreciated :)

What do you get when you differentiate that expression?

 

[Mark44]

So here it says that $d/dt [-\mu/r] = (\mu/r^2)* v$ where $r$ is the separation or radius, and $v$ is the velocity, and $\mu$ is a parameter equal to $G(M+m)$

And I do not see how they get this to be equal, because all of my tries using the quotient rule create another term that they do not have. Any help will be appreciated :)

The formula above implicitly assumes that r is a function of t. $\mu$ is a constant, since the quantities G, M and m are constants, at least as far as the calculation is concerned.
The quotient rule will work here, and is straightforward to use, if you keep in mind that the derivative of a constant is zero. Please show us what you did, and I'm sure we can set you straight.

 

[AdrianGriff]

What do you get when you differentiate that expression?

Thank you guys for your support, I will show what I am doing, as I am surely missing something so plainly obvious..

I did use the Quotient Rule also :)
$$\frac {d} {dx} \left[ \frac {g(x)} {h(x)} \right] = \left( \frac {g'(x)h(x) - h'(x)g(x)} {h(x)^2} \right)$$
And thus:
$$\frac {d} {dt} \left[ \frac {-\mu} {r} \right] = \frac {1\mu^0 * r - (-\mu * r')} {r^2} = \frac {r + \mu * v} {r^2}$$

However, @Mark44 's answer tipped me off and I now realize that $\mu$ itself differentiates to 0, and the first term would cancel to zero as well. So thank you.

 

[Mark44]

Thank you guys for your support, I will show what I am doing, as I am surely missing something so plainly obvious..

I did use the Quotient Rule also :)
$$\frac {d} {dx} \left[ \frac {g(x)} {h(x)} \right] = \left( \frac {g'(x)h(x) - h'(x)g(x)} {h(x)^2} \right)$$
And thus:
$$\frac {d} {dt} \left[ \frac {-\mu} {r} \right] = \frac {1\mu^0 * r - (-\mu * r')} {r^2} = \frac {r + \mu * v} {r^2}$$

I'm pretty sure you get it, from what you wrote below, but here's a comment on your work above. It is not true that $\frac d {dx} \mu = 1\mu^0$. The error here is treating $\mu$ as if it were a function of r. What is correct is that $\frac d {dr} \mu = 0$ -- the derivative of any constant is 0.

However, @Mark44 's answer tipped me off and I now realize that $\mu$ itself differentiates to 0, and the first term would cancel to zero as well. So thank you.

 

[AdrianGriff]

I'm pretty sure you get it, from what you wrote below, but here's a comment on your work above. It is not true that $\frac d {dx} \mu = 1\mu^0$. The error here is treating $\mu$ as if it were a function of r. What is correct is that $\frac d {dr} \mu = 0$ -- the derivative of any constant is 0.

Thank you :)

However the only ambiguity I can feel for here is that is $\mu$ not a function?
If $\mu = G(M+m)$ is that not considered a function during derivation?

 

[Mark44]

Thank you :)

However the only ambiguity I can feel for here is that is $\mu$ not a function?
If $\mu = G(M+m)$ is that not considered a function during derivation?

No. As you said in post #1, $\mu$ is a parameter, a kind of constant. G is the gravitational constant, and M and m are considered constants for the purpose of differentiation. The only varying quantities are r and $\frac \mu r$, that latter of which is a function of r.

 

[AdrianGriff]

No. As you said in post #1, $\mu$ is a parameter, a kind of constant. G is the gravitational constant, and M and m are considered constants for the purpose of differentiation. The only varying quantities are r and $\frac \mu r$, that latter of which is a function of r.

Does treating $\mu$ like a function, and using the chain rule also create the same output? As G, M, and m are all constants and would all derive to be equal to zero? Or must it remain solely as a parameter?

Thank you a lot though :)

 

[Mark44]

Does treating $\mu$ like a function, and using the chain rule also create the same output?

No, if I understand what you're asking. Then again, I don't see how the chain rule is applicable in this problem, since another way of writing the function is $-\mu r^{-1}$, which doesn't call for the chain rule..
In your work you had $\d {dr} \mu = 1\mu^0 = 1$, which is incorrect.

As G, M, and m are all constants and would all derive to be equal to zero?

All would differentiate to zero, yes. (We don't say "derive" when we differentiate something.) You can derive the Quadratic Formula when you solve a quadratic equation by completing the square, but you differentiate $\sin(x)$ to get $\cos(x)$.

 

[PeroK]

Thank you :)

However the only ambiguity I can feel for here is that is $\mu$ not a function?
If $\mu = G(M+m)$ is that not considered a function during derivation?

You can consider it a function of time, but it's a constant function. If you draw a graph of $G(M+m)$ against time, the graph is flat, so the derivative is $0$.

More fundamentally, though, you shouldn't even be trying to differentiate constants like that.