Satellites in Orbit: Mass vs Velocity

[vipson231]

Homework Statement

You can determine the velocity of a satellite in uniform circular orbit by applying Newton's second law, Newton's law of gravity, and the centripetal acceleration due to gravity. Which of the following is not correct?

a) The orbital radius r and the speed v cannot be independently chosen. In other words, for different values of r there is corresponding different values of v.
b) Given the same orbital radius, a satellite in orbit around the Moon will have a slower speed than a satellite in orbit around the Earth.
c) The satellite's speed depends on its mass; the larger the mass, the slower the velocity.
d) Two satelites are in orbit at the same time orbital radius from the planet, One is given a velocity equal to the escape velocity. The escape velocity is is √2 times greater than the velocity of the satelite of the orbit.

I'm pretty sure the answer's C. I can narrow the choices down to b or c. Isn't the speed proprtional to the radius not mass?

 

[haruspex]

I could simply tell you whether C is right, but it will be more helpful to you if you understand why. Please post your reasoning.

 

[vipson231]

I think the answer is based on the formula V= √g x m^2/r hence v is inversely proportional to r not m. Hence the greater the velocity the lesser the mass. So velocity is dependent on radius not mass.

 

[PeterO]

I think the answer is based on the formula V= √g x m^2/r hence v is inversely proportional to r not m. Hence the greater the velocity the lesser the mass. So velocity is dependent on radius not mass.

In your formula, it may show "v is inversely proportional to r", but your formula has an m² in it so that would have v proportional to m in some way as well?

Perhaps your formula is not entirely correct.

Note:
When Astronauts go for a "space walk", they are usually "tethered" for safety reasons. The tether is generally quite loose/limp, meaning it wouldn't matter if it wasn't there [it is there, for safety, in case something unusual happens].

If C was correct, what would happen when a 90kg Astronaut "stepped outside" of the 450000 kg International Space Station?

 

[vipson231]

Good point! Perhaps since it is going in a circular orbit (I'm assuming a tangential direction around the circle) , the speed increases as the radius decreases and vice-versa. I don't think mass would influence an object's speed when going around in a circular orbit. Please correct me if I'm wrong :)

 

[PeterO]

Good point! Perhaps since it is going in a circular orbit (I'm assuming a tangential direction around the circle) , the speed increases as the radius decreases and vice-versa. I don't think mass would influence an object's speed when going around in a circular orbit. Please correct me if I'm wrong :)

Would be good if you could derive the formula which makes no reference to mass of the orbiting object.

Perhaps the formula you had was it - but poorly derived and/or poorly written - as there seemed to be an m term in there. Perhaps that m was not the mass of the orbiting object?? If so, your expression did not make that clear.

 

[vipson231]

The only formula I derived from was the F= Gm1m2/r^2 formula which I then got to be g=m^2/r^2 and eventually V= √Gxm^2/r. However I can't seem to find a simplified version of just speed and radius to prove my point. =/

 

[PeterO]

The only formula I derived from was the F= Gm1m2/r^2 formula which I then got to be g=m^2/r^2 and eventually V= √Gxm^2/r. However I can't seem to find a simplified version of just speed and radius to prove my point. =/

You derived poorly there - or used an incorrect formula.

Try using M and m rather than m₁ and m₂ so you can't possibly mix up which mass is which [I usually use M for the large mass at the centre] you will then see that m₁.m₂ does not equal m² - you would never dream of expressing Mm as m².

What does the x stand for in your V= √Gxm^2/r ?

 

[vipson231]

Sorry for the late reply. The x is multiplication sign.

 

[PeterO]

Sorry for the late reply. The x is multiplication sign.

Would be interested to see your derivation of the expression.

Given the final expression is wrong, there has to be a mistake somewhere.

Note: I don't even like g = m[sup[2[/sup]/r[sup[2[/sup]; but then I am not too sure what the m refers to [Mass of the earth, Mass of the Moon, Mass of the Satellite?]