# Circular Motion Question

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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:

For angular speed I did:

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

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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#2

(Original post by

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:

For angular speed I did:

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

**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 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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#3

**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

i.e.

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#4

(Original post by

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.

**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.

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#5

(Original post by

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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**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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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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#6

**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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i.e.

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#7

(Original post by

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.

**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.

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#8

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#9

(Original post by

Hey how do you use the actual symbols and stuff in your posts ?

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**RossB1702**)Hey how do you use the actual symbols and stuff in your posts ?

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#10

(Original post by

I use LaTeX, it is built in to TSR.

**Cryptokyo**)I use LaTeX, it is built in to TSR.

Thanks.

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#11

**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:

*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 the radius, use Kelper's third law,

Hope it helps.

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#12

(Original post by

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,

Hope it helps.

**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,

Hope it helps.

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