Jump to content

physics help

Featured Replies

"a science group put in a satellite of mass 'm' into a circular orbit of radius 'r'. the orbital velocity it needs to remain in this orbit is 'v'. they now put another satellite into a similiar orbit at the same altitude. its mass is 3 times 'm'. what orbital velocity would it need to be given? give reasons why using mathematical reasoning".

 

ok

i have tried this and have come up with an answer but im not sure i am right.

since it is orbiting i used the equation v^2=Gm/r which is from the fusion of the centripedal and gravitational equations. using this equation 'm' is the mass of the earth as the mass of the satellite is cancelled out, thus i think that the mass would be the same no matter how heavy it is. thus the mass is independant and not used in the equation.

 

can someone please tell me if i am going in the right directio nand please help me to finish this question please!!!!!!

You actually did finish the question already: You get the velocity by setting the centrifugal force F = Mv²/r to equal the attractive gravitational force F = GmM/r². Wheter you replace M by 3M (M is the mass of the satelite; m the mass of earth, here) or not doesn´t play a role. It still cancels out in your equation for v. So the velocity needed to keep a satelite in a stable orbit does not depend on the satelites mass. Given a certain orbit it´s equal for all masses M of satelites.

 

This should not really come to much of a surprise. In fact, a key feature of gravity is that the movement of a body due to a given gravitational field is always independent of that bodies mass.

 

"... i think that the mass would be the same no matter how heavy it is."

^^ That was a typo, wasn´t it?

  • Author

hey, thanks alot man, i really appreciate it, really thanks for double checking, this is just the reassurance i needed.

I´m pretty sure. What do you think is wrong? And why?

It seems illogical that a heavier object needs the same velocity to travel in orbit as a lighter object. I know that gravity acts on all bodies equally, but i'm really having troubles imagining it.

using the formula:

m2*(v^2)/r =G*m1*m2 / (r^2) assuming m2 is the mass of the satellite, it cancels out...

are u sure that means that the mass of the body doesn't affect it's velocity

Yes, I´m pretty sure. You get an equation for the velocity which (directly taken from your post; didn´t check it) is v = sqrt(G*m1/r). Any change in m2 will not alter the value of v. You can also leave m2 in if it makes you feel better: v = sqrt[(G*m1*m2)/(m2*r)]. Still: Changing m2 doesn´t change v (replace m2 with 5*m2, for instance).

 

The mass cancels out at a much more basic level, actually. The force excerted by a gravitational field g on a mass m is F = mg. Now from Newtonian mechanics, you know that F=ma => a = F/m. Hence, the acceleration of a body caused by a gravitational field is the same for all bodies, regardless of their mass: a = F/m = m*g/m = g.

Two bodies which have the same initial positions and the same initial velocities at a given time and do always experience the same acceleration move exactly the same path. So if your heavier satelite starts at the same height and with the same initial velocity as the lighter one, it will have exactly the same orbit.

 

EDIT: Yes, I know it sounds illogical (unintuitive fits better, here). But it´s also counter-intuitive that a feather falls down as fast as a hammer if there´s no friction due to air resistance.

Yes' date=' I´m pretty sure. You get an equation for the velocity which (directly taken from your post; didn´t check it) is v = sqrt(G*m1/r). Any change in m2 will not alter the value of v. You can also leave m2 in if it makes you feel better: v = sqrt[(G*m1*m2)/(m2*r)']. Still: Changing m2 doesn´t change v (replace m2 with 5*m2, for instance).

 

The mass cancels out at a much more basic level, actually. The force excerted by a gravitational field g on a mass m is F = mg. Now from Newtonian mechanics, you know that F=ma => a = F/m. Hence, the acceleration of a body caused by a gravitational field is the same for all bodies, regardless of their mass: a = F/m = m*g/m = g.

Two bodies which have the same initial positions and the same initial velocities at a given time and do always experience the same acceleration move exactly the same path. So if your heavier satelite starts at the same height and with the same initial velocity as the lighter one, it will have exactly the same orbit.

 

EDIT: Yes, I know it sounds illogical (unintuitive fits better, here). But it´s also counter-intuitive that a feather falls down as fast as a hammer if there´s no friction due to air resistance.

 

Thanks buddy

It makes better sense now

I disagree that it is counter-intuitive, Galileo Galilei gave a good intuitive arguement for the idea that two masses should travel in the same path neglecting air resistence. He put forward the thought experiment of having two equal masses separated by a infintesimal thread that you can add or remove, since the presence/absence of the thread is irrelevant then the amount of mass is intuitively irrelevant.

I disagree that it is counter-intuitive, Galileo Galilei gave a good intuitive arguement for the idea that two masses should travel in the same path neglecting air resistence. He put forward the thought experiment of having two equal masses separated by a infintesimal thread that you can add or remove, since the presence/absence of the thread is irrelevant then the amount of mass is intuitively irrelevant.

 

If it's not counterintuitive, then why did everyone think the alternative was true?

If it's not counterintuitive, then why did everyone think the alternative was true?

 

not everyone.... :P

intuition can be useful in getting the correct answer if it is good. then you dont need the maths to back it up, it is much quicker and opperates without pen and paper (or calculator), it is better to educate your intuition as well as learn the formulea, writing off intuition is bad move

not everyone.... :P

 

Up until Galileo, it seems that no one went "on the record" to say that the intuition was wrong. And even today, people have to be taught that it isn't the case.

Up until Galileo, it seems that no one went "on the record" to say that the intuition was wrong. And even today, people have to be taught[/i'] that it isn't the case.

 

and everyone thought the earth was flat, up unitl some dude said otherwise ;)

Not really flat but rather a cube shape. if u went too far, they thought you'd fall off into an abyss.

 

It was the greeks who saw the earth's shadow on the moon and noted that the earth was indeed round. Pretty smart for mediocre technology eh :)

I disagree that people before Galileo did not have a correct intuition about how masses fell. For example if you challenged an ancient to a game of darts with differently massed darts, I don't think they would do significantly worse than someone today.

 

The historical period that Galileo lived in had a dogmatic view about the motion of objects that was specifically Aristotolean, in that view an object fell faster as it neared the ground because the purpose of its fall (to reach the ground) was getting nearer. It was in fact education that Galileo had to overcome not intuition.

Archived

This topic is now archived and is closed to further replies.

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.

Configure browser push notifications

Chrome (Android)
  1. Tap the lock icon next to the address bar.
  2. Tap Permissions → Notifications.
  3. Adjust your preference.
Chrome (Desktop)
  1. Click the padlock icon in the address bar.
  2. Select Site settings.
  3. Find Notifications and adjust your preference.