Using v = sqrt (GM/r)?

[Justin Bieber]

This equation isn't anywhere in the Heinamann textbook nor have I ever heard my teacher mention it.

Is it suitable to use in the VCE exam?

And from my understanding, this can only be used when there's one object moving around the other. Is that correct?
So let's say a 500 kg satellite is in orbit 5 km above the Earth's surface.

Using the equation, the velocity = sqrt ((6.67 * 10^-11 * 500)/(6.37 * 10^6 + 5000))

That gives me 7.2 * 10^-8 (0.000000072) m/s which doesn't seem right at all.


What have I done wrong?

Thank you in advance.

[silverpixeli]

This equation isn't anywhere in the Heinamann textbook nor have I ever heard my teacher mention it.

Is it suitable to use in the VCE exam?

I'm pretty sure it's fine to have this and other pre-derived 'shortcut' formulas on your rules sheets for use in the exam, the trick is to know how to use them correctly so that you don't just plug irrelevant numbers into an irrelevant formula and get an incorrect answer.

Using the equation, the velocity = sqrt ((6.67 * 10^-11 * 500)/(6.37 * 10^6 + 5000))

That gives me 7.2 * 10^-8 (0.000000072) m/s which doesn't seem right at all.

Try using big M, as in, the mass of the earth. Haven't revised motion since before the SAC (should get round to that) but from memory the mass of the object in orbit (small m) doesn't actually affect rotation speed, just like the mass of a brick falling near the earth's surface wont have any effect on acceleration due to gravity. (the whole thing where a feather and a rock fall at the same rate on the moon or in a vacuum in a physics classroom etc.)

[Justin Bieber]

@silverpixeli

Yeah that's so true. Anyone can just plug in numbers on the calculator, but it's about the which ones you use ahaha.


And I used the mass of the Earth instead and it gave me 7900 m/s which seems a lot more right.
And your explanation makes a lot of sense too, so thank you for that!


I haven't touched anything motion related for about 2-3 months now lol, I just started again earlier today.
It feels like I still remember everything, but my firm grasp on all the concepts has turned into a sweaty and slippery grasp. D:

Ah well, these holidays will be a huge catch up session lol.

Thanks for that.


But to everyone else, just to be 100% sure, is all of this correct?

[lzxnl]

Don't use formulas like these. Use physical concepts.
For an object to remain in circular motion, it must receive a net force of mv^2/r directed radially inward. The gravitational force is radially inward, so the direction is satisfied. For the satisfying of the magnitudes:
GMm/r^2 = mv^2/r
Note how now you can clearly identify what each m means; the little m is the mass of the satellite and so doesn't matter.
GM/r^2=v^2/r
v^2=GM/r
as you had earlier. Only now you won't confuse the m's. Other than that, your working is fine.

[Justin Bieber]

Don't use formulas like these. Use physical concepts.
For an object to remain in circular motion, it must receive a net force of mv^2/r directed radially inward. The gravitational force is radially inward, so the direction is satisfied. For the satisfying of the magnitudes:
GMm/r^2 = mv^2/r
Note how now you can clearly identify what each m means; the little m is the mass of the satellite and so doesn't matter.
GM/r^2=v^2/r
v^2=GM/r
as you had earlier. Only now you won't confuse the m's. Other than that, your working is fine.

Oh wow, I just looked at your subject's scores last year and my mind has actually been blown lol.


And well, I'm not really the type of guy to derive equations myself, I'm more of the memorize all of the equations then adapt to the situation type of guy.


But I'm reading your explanation and I understand everything except 'For the satisfying of the magnitudes: GM/r^2=v^2/r'
So when you say 'satisfying of the magnitudes' what do you actually mean by that?

[SocialRhubarb]

I think what he means is that you require a certain force to maintain a certain orbit, and that force must be provided by the gravitational force. Thus, the gravitational force must 'satisfy' the force required for a certain orbit.

This situation is depicted by the equation:



Which of course can be rearranged to form:

[lzxnl]

Think about moving your arms in a circle. You know that if you push harder on your arms, one of three things will happen:
1, increase the speed of rotation
2, decrease the radius of the circle
3, a combination of the above

Also, changing the speed seems to be so much harder than decreasing the radius of the circle. This is because the force mv^2/r, directed radially inward, is a necessary prerequisite for the object to move in a circular path of radius r and velocity v (uniform circular motion just means no acceleration in the direction of the object's movement; that's not needed for this year). If you ever slack off with your arms, either you have to increase the radius or the speed drops. The force mv^2/r is not a force you get; it's what you need.

A circular orbit is an example. If a satellite does not quite have the force to move in that orbit, it will move out further until the gravitational force it receives matches the centripetal acceleration. If you could magically slow the satellite down, that would work too. Don't worry about the possibility of elliptical orbits in VCE. However, if the satellite receives a net force which is directed radially inward, of magnitude mv^2/r and is perpendicular to its motion, it will move in that circle. Hence why I can equate GMm/r^2, the gravitational force, with mv^2/r, for the requirement of uniform circular motion.