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

• February 17th 2011, 10:00 PM
bazingasmile
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,
• February 17th 2011, 10:10 PM
topsquark
Quote:

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
• February 17th 2011, 10:19 PM
topsquark
Quote:

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
• February 17th 2011, 10:23 PM
bazingasmile
Thank you
Thank you very much.
I'm not stuck anymore!
Thanks for also pointing out my typo.

G.
• February 18th 2011, 05:44 AM
topsquark
Quote:

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. (Sun)

-Dan