You are Here: Home >< Physics

# OCR Physics G484 - June 2013 Unit 4 (OFFICIAL RETAKE THREAD) Watch

1. (Original post by eggfriedrice)
Q) in order to move a satellite into a new smaller orbit, a decelerating force is applied for a brief period of time. Suggest how the decelerating for could be applied.

Now I don't understand the question. We want a smaller orbit so smaller r. Surely by decelerating the orbit or decreasing the force, wouldn't the radius increase? The mark scheme says to apply something like a rocket motor against the direction of travel, thus decelerating it.

Can anyone explain this to me? Thanks.
I see your logic but the answer to the question is to simply use a rocket in opposite direction, a smaller radius at the same velocity the centripetal force outwards would be too great the the satellite would fling off into space and too low the centripetal force outwards won't be great enough and the satellite would fall towards Earth
2. (Original post by leather0294)
finally got my hands on them
Thank you very much!

(Original post by Gulzar)
But u must remember that the thermodynamic scale is made in such a way that a change in temp of say 5 degrees celsius is the SAME as the temp change of 5 Kelvin
Oops, I forgot about that, thanks for reminding me.
3. (Original post by martynsteel)
I see your logic but the answer to the question is to simply use a rocket in opposite direction, a smaller radius at the same velocity the centripetal force outwards would be too great the the satellite would fling off into space and too low the centripetal force outwards won't be great enough and the satellite would fall towards Earth
I still don't understand, a rocket to oppose the direction of velocity or the net force? Either way wouldn't it push the satellite out and into a bigger orbit.
4. Any guesses on what might come up based on previous papers??
5. (Original post by eggfriedrice)
I still don't understand, a rocket to oppose the direction of velocity or the net force? Either way wouldn't it push the satellite out and into a bigger orbit.
I think it helps if you think in terms of F=mv^2/r, hence r=mv^2/F

You want a smaller radius, so since r is proportional to v^2, if you decrease v, then r will decrease.

By using a rocket to oppose the motion, you are effectively reducing the velocity of the satellite, since the net force in the direction of motion of the satellite is reduced.

Hope this makes sense/is correct

Posted from TSR Mobile
6. (Original post by imogen--)
Any guesses on what might come up based on previous papers??
I'm not sure, but I guess if you check the specification against Jan 2013, June 2012 and onwards if you want to, you can cross off things that have come up and that will leave things that haven't come up recently (or at all if you looked at every paper).

Posted from TSR Mobile
7. Does anyone know what UMS you need to get an A* in physics?
8. (Original post by Jullith)
Does anyone know what UMS you need to get an A* in physics?
480/600 UMS in total, 270 UMS of which are from A2 units.

Posted from TSR Mobile
9. (Original post by Rhodopsin94)
I think it helps if you think in terms of F=mv^2/r, hence r=mv^2/F

You want a smaller radius, so since r is proportional to v^2, if you decrease v, then r will decrease.

By using a rocket to oppose the motion, you are effectively reducing the velocity of the satellite, since the net force in the direction of motion of the satellite is reduced.

Hope this makes sense/is correct

Posted from TSR Mobile
It does make sense in that sense, but at the same time F would decrease so r would increase. Unless the velocity has a greater affect than the force...

Actually now that I think about it, decreasing the velocity WOULD have a greater affect since r is proportional to v^2.
10. (Original post by eggfriedrice)
It does make sense in that sense, but at the same time F would decrease so r would increase. Unless the velocity has a greater affect than the force...

Actually now that I think about it, decreasing the velocity WOULD have a greater affect since r is proportional to v^2.
I see your point about v^2 having a greater effect than force, but you have to remember that the decelerating force has nothing to do with the centripetal force (directly) - they are separate forces.

You know that the direction of velocity is always perpendicular to the centripetal force (and acceleration) of a body in circular motion. Therefore the decelerating force has no component in the same direction as the centripetal force. You need to remember that the F in F=mv^2/r is centripetal force.

The decelerating force still reduces the velocity etc, but is not part of the F=mv^2/r equation.

So I think your main confusion with this has not been distinguishing between the centripetal force and external forces which are at right angles to the centripetal force. Hopefully it all makes sense now though

Posted from TSR Mobile
11. (Original post by Rhodopsin94)
I see your point about v^2 having a greater effect than force, but you have to remember that the decelerating force has nothing to do with the centripetal force (directly) - they are separate forces.

You know that the direction of velocity is always perpendicular to the centripetal force (and acceleration) of a body in circular motion. Therefore the decelerating force has no component in the same direction as the centripetal force. You need to remember that the F in F=mv^2/r is centripetal force.

The decelerating force still reduces the velocity etc, but is not part of the F=mv^2/r equation.

So I think your main confusion with this has not been distinguishing between the centripetal force and external forces which are at right angles to the centripetal force. Hopefully it all makes sense now though

Posted from TSR Mobile
Ahh I get it. F=mv^2/r v decreases but F doesn't because the decelerating force is acting perpendicular to it. Thanks again!
12. Guys predictions of any experiment questions, definitions or deriving?
13. Anyone got a list of the experiments we should memorise? Looked through spec but couldn't see specific mentions of any experiments so I'm not sure what we need to know - I think I did see specific heat capacity though but I don't know the method for it.
14. (Original post by Rhodopsin94)
check the specification against Jan 2013, June 2012 and onwards if you want to, you can cross off things that have come up and that will leave things that haven't come up recently
I have just looked at the specification against January 2013, and based on that the topics that weren't mentioned at all or very much are:
-Circular motion
-SHM Graphs
-Resonance
-Kinetic Theory of Gases/Brownian Motion

This could mean a Brownian motion experiment long answer question might come up, does anyone have a good summary to answer that???
15. for kepler's 3rd law, markscheme says:

The cube of the planets distance (from the Sun) divided by the square of the
(orbital) period is the same (for all planets) (WTTE)

would it be find to say this the other way around? square of period/cube of distance? i believe that would give the same answer
16. When do we use a=v^2/r?
17. I understand everything pretty well apart from these ridiculously simple questions, how do you estimate the area under a curve like this?

18. (Original post by imogen--)
I understand everything pretty well apart from these ridiculously simple questions, how do you estimate the area under a curve like this?

Draw a triangle and work out the area.
Then you need to count the squares that weren't included in the area and add them to the area of the triangle.
You can do this by finding out the area for one box lxw.
Be careful of the units for time, here they are in ms.
Draw a triangle and work out the area.
I'm really sorry I am being stupid here but I still don't really understand, where do you draw the triangle? When I have been attempting to do what you said i have drawn a right angled triangle inside the curve with the highest point at the top of the curve, is that right? If it is, it's really hard to count the squares accurately because it crosses over lots of squares. Sorry I don't know why this is confusing me so much x

Do you only count the whole squares or something?
20. (Original post by imogen--)
I'm really sorry I am being stupid here but I still don't really understand, where do you draw the triangle? When I have been attempting to do what you said i have drawn a right angled triangle inside the curve with the highest point at the top of the curve, is that right? If it is, it's really hard to count the squares accurately because it crosses over lots of squares. Sorry I don't know why this is confusing me so much x

Do you only count the whole squares or something?
I just count the number of squares, mark each box with a number. Add the "partial" squares to make 1 square until you have covered all the area under the graph. Multiply the number of squares by the area of 1 square. The mark scheme allows for some error in square counting.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: December 13, 2013
Today on TSR

### Does race play a role in Oxbridge admissions?

Or does it play no part?

### The worst opinion about food you'll ever hear

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources

Can you help? Study help unanswered threadsStudy Help rules and posting guidelinesLaTex guide for writing equations on TSR

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.