why the mass of the planet plays no role in determining a satellite's orbital speed (1 Viewer)

malcolm21 · Oct 24, 2015

I thought the formula was root 2 G mass of planet / r

malcolm21 · Oct 24, 2015

Re: why the mass of the planet plays no role in determining a satellite's orbital spe

omg i fucked up sorry it was the planet orbitting something else lol

forget about this question

Physicklad · Oct 24, 2015

Re: why the mass of the planet plays no role in determining a satellite's orbital spe

theres no 2

just (GM/r)^1/2

Mr_Kap · Oct 24, 2015

Re: why the mass of the planet plays no role in determining a satellite's orbital spe

malcolm21 said:
I thought the formula was root 2 G mass of planet / r

its Squar3root(GM/r)

Squar3root · Oct 24, 2015

Re: why the mass of the planet plays no role in determining a satellite's orbital spe

isn't that escape velocity? lol

malcolm21 · Oct 24, 2015

Re: why the mass of the planet plays no role in determining a satellite's orbital spe

lol oops

InteGrand · Oct 24, 2015

Re: why the mass of the planet plays no role in determining a satellite's orbital spe

It's not the mass of the planet that plays no role in determining the satellite's orbital speed, it's the mass of the satellite.

Crisium · Oct 24, 2015

Re: why the mass of the planet plays no role in determining a satellite's orbital spe

Orbital Velocity = GM / r (square root over the whole thing)

Its derivation comes from equating the centripetal force formula and the gravitational force equation that includes the mass of the planet and the mass of the object (The mass of the object is cancelled out in the derivation hence why it isn't dependent on the mass of the object).

Escape Velocity = 2GM / r (square root over the whole thing)

It's derivation comes from equation the kinetic energy formula and the gravitational potential energy formula

InteGrand · Oct 24, 2015

Re: why the mass of the planet plays no role in determining a satellite's orbital spe

Crisium said:
Escape Velocity = 2GM / r (square root over the whole thing)

It's derivation comes from equation the kinetic energy formula and gravitational force equation that includes the mass of the planet and the mass of the object (The mass of the object is cancelled out in the derivation hence why it isn't dependent on the mass of the object).

Actually gravitational potential energy.

Crisium · Oct 24, 2015

Re: why the mass of the planet plays no role in determining a satellite's orbital spe

InteGrand said:
Actually gravitational potential energy.

Yeah my bad haha

What happens to the negative though?

InteGrand · Oct 24, 2015

Re: why the mass of the planet plays no role in determining a satellite's orbital spe

Crisium said:
Yeah my bad haha

What happens to the negative though?

Well actually, we don't equate the kinetic energy to the potential energy (if we did, there'd be a negative, as you said). What we actually do is equate the total mechanical energy to 0. It is explained here:

InteGrand said:
Escape velocity is the velocity $v_\text{esc.}$ given by $E_k + E_p = 0\Leftrightarrow\frac{1}{2}mv_\text{esc.}^2+\left ( -\frac{GmM}{R_\text{planet}} \right )=0\Leftrightarrow v_\text{esc.}=\sqrt{\frac{2GM}{R_\text{planet}}}$ , where M is the mass of the planet, and R_planet is the radius of the planet. Thus as you can see, the square root of the gravitational constant helps determine v_esc., and v_esc. is proportional to the square root of the mass of the planet, and inversely proportional to the square root of the planet's radius.

Here is the explanation why (i.e. how we derived the formula for v_esc.).

We are assuming there's no loss of energy from the spacecraft due to air resistance or other frictional effects etc. Therefore, the total mechanical energy (which is E_tot. = E_k + E_p) of the craft is conserved throughout its flight (conservation of energy). This means the value of E_tot. = E_k + E_p is constant at any point in time of the craft's flight. As we are launching it at escape velocity, we want in the limit at time goes to infinity, the distance the craft is from the planet to tend to infinity (this is the definition of the escape velocity: the craft will get arbitrarily far away from the planet as time goes on and on – it doesn't reach a maximum distance and then fall back down to Earth), and in this limit, the speed of the spacecraft will tend to 0. This means in the limit as $t\to \infty$ , we have $E_k\to0$ (because the velocity is tending to 0 as the earth's gravity keeps slowing the craft down) and $E_p\to0$ (because the distance r of the craft tends to infinity, and as r is in the denominator of the E_p formula, $E_p\to0$ ). This means the sum of kinetic energy and potential energy (i.e. total mechanical energy), in the limit as $t\to0$ , is 0.

Since $\lim_{t\to\infty}E_\text{tot.}(t)=0$ , and we said earlier that by conservation of energy, $E_\text{tot.}(t)$ is constant, it means this constant value is 0, i.e. $E_\text{tot.}(t)=0$ $\forall t\geq0$ .

$\therefore E_\text{tot.}(0)=0\Rightarrow\frac{1}{2}mv_\text{esc.}^2+\left ( -\frac{GmM}{R_\text{planet}} \right )=0$ , since when time = 0, the craft starts at the planet's surface (hence E_p has R_planet in the denominator, as this is the initial distance from the planet's centre), and the velocity is the escape velocity (since we're launching it at escape velocity), so E_k initially equals $\frac{1}{2}mv_\text{esc.}^2$ (m is the mass of the craft).

So that's the explanation of how we got the equation at the start, from which we derived escape velocity's formula.

Crisium · Oct 24, 2015

Re: why the mass of the planet plays no role in determining a satellite's orbital spe

InteGrand said:
Well actually, we don't equate the kinetic energy to the potential energy (if we did, there'd be a negative, as you said). What we actually do is equate the total mechanical energy to 0. It is explained here:

Yeah I had a feeling equating them would make no sense

Thanks for reminding me

(I haven't done calculation questions involving this in a while)

InteGrand · Oct 24, 2015

Re: why the mass of the planet plays no role in determining a satellite's orbital spe

$Alternatively, it is can be derived via calculus. Let $x=x(t)$ be the displacement of the satellite from the planet's centre at time $t\geq 0$. We know from Newton that $\ddot{x}= -\frac{GM}{x^2}$,where $M$ is the planet's mass and dots represent time derivatives. From HSC 3U mechanics, we have$

$v\frac{\mathrm{d}v}{\mathrm{d}x}= -\frac{GM}{x^2}$

$$\Rightarrow \int _{v_{\text{esc.}}} ^{0} v\text{ d}v = \int_{R} ^{\infty} -\frac{GM}{x^2} \text{ d}x$ (rearranging and integrating; the reason for the limits of integration is that at the lower limit, when $t=0$, if the object is launched from $x=R$ (i.e. surface of planet of radius $R$), at a speed of $v_{\text{esc.}}$ (escape velocity), then by definition of escape velocity, as $t\to \infty$, we have $x\to \infty$ and $v\to 0$)$

$$\Rightarrow \Big{[}\frac{v^2}{2} \Big{]}_{v=v_{\text{esc.}}} ^{v=0} = \Big{[}\frac{GM}{x} \Big{]}_{x=R} ^{x=\infty}$ (basic integration)$

$$\Rightarrow 0-\frac{v_{\text{esc.}} ^2}{2} = 0 - \frac{GM}{R}$ (as $\lim _{x\to \infty} \frac{GM}{x}=0$)$

$\Rightarrow \frac{v_{\text{esc.}} ^2}{2} = \frac{GM}{R}$

$$\Rightarrow \boxed{v_{\text{esc.}} = \sqrt{\frac{2GM}{R}}}$.$

why the mass of the planet plays no role in determining a satellite's orbital speed (1 Viewer)

malcolm21

Active Member

malcolm21

Active Member

Physicklad

Member

Mr_Kap

Well-Known Member

Squar3root

realest nigga

malcolm21

Active Member

InteGrand

Well-Known Member

Crisium

Pew Pew

InteGrand

Well-Known Member

Crisium

Pew Pew

InteGrand

Well-Known Member

Crisium

Pew Pew

InteGrand

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)