Space (1 Viewer)

Kaido · Jul 24, 2015

GPE = Work - Kinetic energy

Can someone verify if this formula is true for a satellite put into orbit, if so, how is this derived?

mrpotatoed · Jul 24, 2015

isn't it just change in kinetic energy = change in gpe, and also doesnt work done = gpe?

since W=Fs=mgh=Ep (subbing the gravity formula for g)

InteGrand · Jul 24, 2015

To get it into orbit, you need to change its kinetic energy (give it kinetic energy sufficient to keep it in orbit at the distance r from the Earth's centre), and also do work against gravity to get it up there, so the total work done would be the sum of the work done against gravity to get it to its orbital radius, and the work done to increase its kinetic energy from 0 (which is its KE at ground level) to the required KE for orbit.

dan964 · Jul 24, 2015

^^ that is correct
W=mas (using F=ma)
s is height (which we can also denote as h)
a is gravity (which equals GM/r^2 or g)
therefore W=mgh or the other formula

The total work is the KE + GE as Integrand has already mentioned.

InteGrand · Jul 24, 2015

mrpotatoed said:
isn't it just change in kinetic energy = change in gpe, and also doesnt work done = gpe?

since W=Fs=mgh=Ep (subbing the gravity formula for g)

Work done to get it up into the orbital radius would not be 'mgh', because we can't assume g is constant for larger r (technically it isn't constant for small r either, but its variation is negligible for small r, which is why we assume its constant for things like projectile motion, as it greatly simplifies calculations and produces answers that are suitable for any practical purpose).

If we (the spacecraft) start at ${r_E}$ (Earth radius), and go to some radius of orbit ${r}$ , where r > r_E, then the work done against gravity to go there is given by the following integral:

$$W_{\text{against gravity}}=\left | \int _{r_E} ^{r} \bold{F}(x)\text{ d}x\right |$, where $\bold{F}(x)$ is the gravitational force when you are $x$ units from the Earth's centre)$

$\Rightarrow W_g=\left | \int _{r_E} ^{r} -\frac{GmM}{x^2}\text{ d}x\right |$, using the symbol $W_g$ to represent the work done against gravity, where $M$ is the Earth's mass and $m$ is the spacecraft's mass, and $m\ll M$

$=\left | \left[\frac{GmM}{x} \right]_{r_E} ^{r} \right |$

$\Rightarrow W_g (r)= GmM\left(\frac{1}{r_E}-\frac{1}{r} \right)$

$$We can see that as $r\to \infty$, $W_g (r) \to \frac{GmM}{r_E}$, which is the $GPE$ at the Earth's radius, which is to be expected, as this is how $GPE$ is defined (the work done by gravity to move you from an infinitely far away point to a given distance $r$ from earth)$.$

$Also, at distance $r$, the required orbital velocity is $v=\sqrt{\frac{GM}{r}}$ (found by equating centripetal force $\frac{mv^2}{r}$ with gravitational force $\frac{GmM}{r^2}$ and solving for $v$), so $v^2 = \frac{GM}{r}$. Hence the required $KE$ for orbit at $r$ is $\frac{1}{2}mv^2 =\frac{GmM}{2r}$. This work must be done on the spacecraft to give it the right velocity for orbit (assuming it starts at rest). So the total work done, $W_T$, is$

$W_T (r) = GmM\left(\frac{1}{r_E}-\frac{1}{r} \right) + \frac{GmM}{2r}$

$$\Rightarrow W_T (r) = GmM\left(\frac{1}{r_E}-\frac{1}{2r} \right)$.$

Space (1 Viewer)

Kaido

be.

mrpotatoed

Active Member

InteGrand

Well-Known Member

dan964

what

InteGrand

Well-Known Member

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