Quick question

[positron]

Umm

 

I got this question but dont really know what to do with it.

 

==> Astronauts in a space capsule orbiting the earth at a height of 900km experience weightlessness: hence the gravitational field due to the earth must be zero at that altitude.

 

It says to include calculations in my answer but i dont know what calculations would be involved.

 

i'm guessing that the statement is wrong and must be proved so by using calculations. can u help me?

[timo]

a) very bad title for your thread

b) how´s "gravitational field" defined in your class?

c) gravitational force is given by F = G*m1*m2/d² with d=900 km, m1 being mass of the earth and m2 being mass of the astronaut. Neither m1 nor m2 nor G is zero. Hence, F is non-zero. If the answer to b) was "there is no force due to gravity", then you´ve got your answer.

d) although it isn´t asked for (teachers tend to demand answers that aren´t asked for explicitely): The astronaut experiences weightlessness because gravitational force and centrifugal force cancel out.

[positron]

Thanks, i was in a rush when i wrote the question and 'quick question' was the only thing that came into my head

[Moebius]

There is a formula to calculate this

 

F (centripetal) = m2*(v^2)/r (pulls you outwards)

F (gravity) = G*m1*m2 / (r^2) (pulls you earthwards)

 

m1 = earth mass

m2 = spaceship mass (cancels, look the formula, so is independent)

v = spaceship velocity

G = Constant

r = distance to the earth center (radius)

 

So when

 

F(centripetal) = F (gravity)

 

m2*(v^2)/r =G*m1*m2 / (r^2)

 

The force you are receiving is 0.

 

Using this formula you must calculate what speed you need to be in orbit, at that speed forces will cancel out and you will feel weightless and spaceship will keep at the same altitude (will not fall) .

 

If you are using a "wrong" speed, one which does not make forces cancel out, you will start to feel the gravity and you will fall to earth.

[JoeMK]

We gotta do sum calculations, but i don't think they gotta be that complex. Just a good theory behind a logical explanation

[JoeMK]

There is a formula to calculate this

 

F (centripetal) = m2*(v^2)/r (pulls you outwards)

F (gravity) = G*m1*m2 / (r^2) (pulls you earthwards)

 

m1 = earth mass

m2 = spaceship mass (cancels' date=' look the formula, so is independent)

v = spaceship velocity

G = Constant

r = distance to the earth center (radius)

 

So when

 

F(centripetal) = F (gravity)

 

m2*(v^2)/r =G*m1*m2 / (r^2)

 

The force you are receiving is 0.

 

Using this formula you must calculate what speed you need to be in orbit, at that speed forces will cancel out and you will feel weightless and spaceship will keep at the same altitude (will not fall) .

 

If you are using a "wrong" speed, one which does not make forces cancel out, you will start to feel the gravity and you will fall to earth.[/quote']

 

you said that "F (centripetal) = m2*(v^2)/r (pulls you outwards)"

Centripetal force pulls you inwards, then u must mean centrifugal force... however, u put the formula for centripetal, which must be wrong

[swansont]

you said that "F (centripetal) = m2*(v^2)/r (pulls you outwards)"

Centripetal force pulls you inwards' date=' then u must mean centrifugal force... however, u put the formula for centripetal, which must be wrong[/quote']

 

It's the same equation, just in a different frame of reference.

 

You move in a circular orbit if the centripetal force is equal to the gravitational force.

 

In the moving frame, there is a pseudoforce (centrifugal) which makes the net force seem to be zero. So it must have the same magnitude as the gravitational force.

 

It's the same solution, just looked at two different ways. Since they describe the same phenomenon, they'd better end up with the same answer.

[gnpatterson]

I expect you to be hung, drawn and quartered if you dare to mention a fictious force that cancels gravity.

 

The concept is rightly being expunged from the physics texts of all civilisation. The concept of "freefall" and "apparent weightless" are much more useful. The current dogma is that space craft in _any_ orbit be it circular, elliptical or hyperbolic are moving in freefall therefore the occupants experience no weight relative to the craft.

 

You should only be expected by your teacher to realise that the gravitational field at 900 km up is in the first approximation the same as that on the surface. If you need to calculate it I strongly suggest that you use the radius of the earth as 6400 km and the radius for your craft as 7300 km = 6400 + 900. You should first show that you can calculate g using the text book values for G,R,M the gravitational constant, earth radius and earth mass.

 

M=5.9736×10^24 kg

G=6.674×10^-11 Nm2kg-2

R=6.356×10^6 m

 

using g(surface)=GM/(r*r) = appox 10 N/kg

 

at orbital height g(orbit) = approx 7.6 N/kg

 

the explanation you must give/understand is that weightlessness would be experienced by any spacecraft in free fall at that point (and any other point) in space because the craft is in free fall. If the craft is in freefall orbit, if the craft has been shot out of a large cannon and is at the top of its arc, if the craft was suspended by a piece of string from the moon that was cut just a second ago and is now falling toward earth, if the craft was shot from jupiter and is headed straight for the sun, it is all irrelevant to the experience of weightless. (of course the people inside might care about which of the different scenarios is involved, the phyics does not)