# Math Help - Differential equation solving related to 2nd Newton's law

1. ## Differential equation solving related to 2nd Newton's law

Without taking into account friction and resistance to the air, but while taking into account the gravitation force and the altitude. The weight $W$ is decreasing in terms of $x$ (altitude):
$P(x)=\dfrac{mgR^2}{(R+x)^2}$
$m$: Mass of the object
$g$: Gravitational force (at sea level)
$R$: Radius of Earth
$x$: Altitude

What is the speed of the object in terms of its altitude (we're looking for $v(x)$)?

I'm trying hard to resolve this equation but just can't find how to relate it to the altitude instead of the time.

Thanks to anyone who takes the time to answer this question. Good night,

2. Originally Posted by bazingasmile
$P(x)=\dfrac{mgR^2}{R+x^2}$
A slight change in the forumla (I presume this is a typo.)
$\displaystyle P(x)=\dfrac{mgR^2}{(R+x)^2}$

-Dan

3. Originally Posted by bazingasmile
Without taking into account friction and resistance to the air, but while taking into account the gravitation force and the altitude. The weight $W$ is decreasing in terms of $x$ (altitude):
$P(x)=\dfrac{mgR^2}{(R+x)^2}$
$m$: Mass of the object
$g$: Gravitational force (at sea level)
$R$: Radius of Earth
$x$: Altitude

What is the speed of the object in terms of its altitude (we're looking for $v(x)$)?

I'm trying hard to resolve this equation but just can't find how to relate it to the altitude instead of the time.

Thanks to anyone who takes the time to answer this question. Good night,
$\displaystyle F = ma \implies a = \frac{gR^2}{(R + x)^2}$

Now we know that a = dv/dt, so
$\displaystyle \frac{dv}{dt} = \frac{gR^2}{(R + x)^2}$

And using the chain rule:
$\displaystyle \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = \frac{dv}{dx} \cdot v(x)$

So we have the differential equation:
$\displaystyle v(x) \cdot \frac{dv}{dx} - \frac{gR^2}{(R + x)^2} = 0$

Can you take it from here?

-Dan

4. ## Thank you

Thank you very much.
I'm not stuck anymore!
Thanks for also pointing out my typo.

G.

5. Originally Posted by bazingasmile
Thank you very much.
I'm not stuck anymore!
Thanks for also pointing out my typo.

G.
Always happy to help out.

-Dan