Weight force and jupiter (1 Viewer)

velox

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There's a question that asks you to find the weight force of an astronaut who weighs 80kg on different planets. Thats easy. But then it asks to "Suggest a reason why we did not ask you for the weight force or Andy on the surface of jupiter"

Why is this? The answers say "They are gas giants. The only solid surface is deep within the gas ball." What does this mean? The acceleration due to gravity is huge on the gas giants eg Jupiter with ~24ms^2
 

Xayma

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I am assuming it is because you werent given the mass/radius of the solid surface.
 

velox

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we are given the mass x Earth and also the Radius x Earth.
And of course the acceleration due to gravity. Oh ic, because the gas giants are mostly gas, and the solid cores are only deep within the planet?
 

helper

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So you've come across Marsden's warped and irrelelvent sense of humour. Question is pointless and not needed for the HSC.
The radius of Jupiter that is quoted is for the radius of the gas giant. When we talk about surface of an object we are normally talking about a solid or liquid boundary, that is seperate from the atmosphere. There isn't a surface as such, so you wouldn't ask for the weight force at the "surface". Yes you could ask for the weight force at that point but not the surface.
 

velox

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Next Question -
Andy (mass 80kg) likes to visit an asteroid called Beowulf. When he stands on Beowulf looking at the stars with his gf Mandy (mass 60kg), he feels a downward force of 5.0N. Find:

d) Mandy’s twin sister rockets towards the asteroid using her jet belt. Calculate the gravity force acting on Sandy when 10km out from the ground.

I cant seem to get the right answer. My answer is way too big.
 

Xayma

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Umm you need a distance from the centre of the asteriod to the surface :confused:
 

helper

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For Andy
F=Gm1m2/d^2
5=GM × 80 /10 000^2
GM = 6 250 000


For twin
F= 6 250 000 × 60/ 20 000^2
= 0.9 N

Remember to square d and to convert to metres

Don't guarantee calculations but method correct, assuming radius was correct
 

Jase

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HAHAHA. I did laugh, and I did cry.
But whoever wrote that question should die.
 

clonestar

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To work out the part d:

F=G m1 * m2/rsquared


so = 6.67 * 10 -11 * 60kg * 9.37 * 10 power of 16/ 20000 squared


=


0.94N



Hope this helps

I am teaching my students all o this as well.. :)

Enjoy

CloNEster
 

velox

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Hmm I now have some problems on understanding the concept of negative gravitational potential energy. Some questions have the negative sign, some dont?? Am i missing something here? How can an object have a negative gpe, if gpe is the like, the higher the object the more gpe it has, and the lcloser it is to the mass, the less gpe it has.

If an object hits earth, wouldn't it have zero gpe when it is lying on the ground? How can you have negative gpe??

Thanks for any help:)
 

helper

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wrx said:
Hmm I now have some problems on understanding the concept of negative gravitational potential energy. Some questions have the negative sign, some dont?? Am i missing something here? How can an object have a negative gpe, if gpe is the like, the higher the object the more gpe it has, and the lcloser it is to the mass, the less gpe it has.
All the above is correct.

An object is going to have GPE when it is in a gravitational field. So the further you are away from the object the greater the GPE.
The only point without a gravitational field is at a distance of infinity. This point therefore will have absolute zero potential energy. Any point closer than this must have less GPE than zero as you are moving closer to the object so less GPE. The only way this can be is if the GPE is negative.

You can also explain it in terms of work. If you don't understand above. I can try that.

If an object hits earth, wouldn't it have zero gpe when it is lying on the ground? How can you have negative gpe??
What is your reference?
What you wrote here is the traditional reference. However, if you dug a hole under the object it would then have potential energy again because it is going to fall to the bottom of the hole. So now the bottom of the hole is zero. So your original point couldn't have been zero. This can keep going. This is why they define zero as at an infinite distance from Earth, where the gravitational force is zero.

ps. If you have a question ask. I didn't substitute G in the other point to reduce the need for exponentials. A lot of students type exponentials wrong on their calculators and because GM was a constant, it was as easy to find GM as just M by itself.
 

velox

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Thanks for clearing that up, but i still have some uncertainties about this topic. I've been looking at marsdens notes, but i just realised that none of the questions from that section demonstrate gpe that well. As none of them hae negative signs....if you get what i mean..Cheers, thanks for ur help again helper:)
Btw, would you say marsdens notes are decent or...? Just wondering if i should be doing textbook questions also...but i haven't got my textbook yet:(
 
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helper

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If you use the E<sub>p</sub>=-Gm<sub>1</sub>m<sub>2</sub>/r formula it should always be quoted as a negative.

It will be a positive if they ask for a change in PE. Eg What is the gain in PE when changing from an Altitude of 100km to 200km.

It wouldn't surprise me if this is in this years HSC as it hasn't been tested yet.
 
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