Blake Jones
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#1
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Hi guys,

I can't see how I'm meant to work this question out, it's like we're missing a bit of information.

The question says:

The Hubble space telescope was launched in 1990 into a circular orbit near Earth. It travels around the Earth once every 97 minutes.
Calculate the angular speed.
Calculate the radius of orbit.

For angular speed I did:

97mins = 5820 seconds
2Pi / 5820 = 1.08x10^-3 rads^-1

But for the radius I can't work out what I'm meant to do. Like I said I feel like I'm missing info.

Thanks
Blake
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username2548927
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(Original post by Blake Jones)
Hi guys,

I can't see how I'm meant to work this question out, it's like we're missing a bit of information.

The question says:

The Hubble space telescope was launched in 1990 into a circular orbit near Earth. It travels around the Earth once every 97 minutes.
Calculate the angular speed.
Calculate the radius of orbit.

For angular speed I did:

97mins = 5820 seconds
2Pi / 5820 = 1.08x10^-3 rads^-1

But for the radius I can't work out what I'm meant to do. Like I said I feel like I'm missing info.

Thanks
Blake
Part a is completely right.
Part B I can't seem to figure it out either, you would think they would give you the tangential velocity so you could do v=rw or give you the circumference of orbit so you could do feta=s/r but i can't see how you're supposed to do it with the data given.


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Cryptokyo
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(Original post by Blake Jones)
Hi guys,

I can't see how I'm meant to work this question out, it's like we're missing a bit of information.

The question says:

The Hubble space telescope was launched in 1990 into a circular orbit near Earth. It travels around the Earth once every 97 minutes.
Calculate the angular speed.
Calculate the radius of orbit.

For angular speed I did:

97mins = 5820 seconds
2Pi / 5820 = 1.08x10^-3 rads^-1

But for the radius I can't work out what I'm meant to do. Like I said I feel like I'm missing info.

Thanks
Blake
For part b. Use the fact that its gravitational attraction to earth must equal the outwards force from the circular motion as it remains in orbit.

i.e. mg=\frac{mv^{2}}{r}=mr\omega^{2}
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username2548927
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(Original post by Cryptokyo)
For part b. Use the fact that its gravitational attraction to earth must equal the outwards force from the circular motion as it remains in orbit.

i.e. mg=\frac{mv^{2}}{r}=mr\omega^{2}
So you're setting the gravitational attraction equal to the centripetal force. I get the logic in it but it doesn't give you the mass of the satellite.


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Joinedup
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(Original post by RossB1702)
So you're setting the gravitational attraction equal to the centripetal force. I get the logic in it but it doesn't give you the mass of the satellite.


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You don't need the mass of the satellite providing it's much less than the mass of the body it's orbiting.

A familiar analogy would be that apart from the effect of air resistance a hammer and a feather fall at the same rate of acceleration in a gravitational field.

So a beercan and a battleship would orbit earth with the same period if they were both in circular orbits of the same altitude.
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Cryptokyo
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(Original post by RossB1702)
So you're setting the gravitational attraction equal to the centripetal force. I get the logic in it but it doesn't give you the mass of the satellite.


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The mass would cancel

i.e. g=r\omega^{2}

\therefore r=\frac{g}{\omega^{2}}
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username2548927
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(Original post by Joinedup)
You don't need the mass of the satellite providing it's much less than the mass of the body it's orbiting.

A familiar analogy would be that apart from the effect of air resistance a hammer and a feather fall at the same rate of acceleration in a gravitational field.

So a beercan and a battleship would orbit earth with the same period if they were both in circular orbits of the same altitude.
Thanks


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username2548927
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(Original post by Cryptokyo)
The mass would cancel

i.e. g=r\omega^{2}

\therefore r=\frac{g}{\omega^{2}}
Hey how do you use the actual symbols and stuff in your posts ?


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Cryptokyo
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(Original post by RossB1702)
Hey how do you use the actual symbols and stuff in your posts ?


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I use LaTeX, it is built in to TSR.
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username2548927
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(Original post by Cryptokyo)
I use LaTeX, it is built in to TSR.
Is there a tutorial on how to use them ? Can you give me a link if so ?

Thanks.


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Eimmanuel
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#11
(Original post by Blake Jones)
Hi guys,

I can't see how I'm meant to work this question out, it's like we're missing a bit of information.

The question says:

The Hubble space telescope was launched in 1990 into a circular orbit near Earth. It travels around the Earth once every 97 minutes.
Calculate the angular speed.
Calculate the radius of orbit.

For angular speed I did:

97mins = 5820 seconds
2Pi / 5820 = 1.08x10^-3 rads^-1

But for the radius I can't work out what I'm meant to do. Like I said I feel like I'm missing info.

Thanks
Blake
The missing info would be the mass of the Earth which I think it is given in the data page of every physics examination paper.

For the radius, use Kelper's third law,

 T^2 = \frac{4 \pi^2}{G M_E} R^3

Hope it helps.
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Cryptokyo
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(Original post by Eimmanuel)
The missing info would be the mass of the Earth which I think it is given in the data page of every physics examination paper.

For the radius, use Kelper's third law,

 T^2 = \frac{4 \pi^2}{G M_E} R^3

Hope it helps.
Kepler's Law is just derived from what I am saying I.e. \frac{GMm}{r^{2}}=mr\omega^{2}


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Blake Jones
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Ah I see, thank you all!
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Nahilachowdhury
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I think it’s the mass of the Earth
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