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# Gravitational Fields questions (Multiple choice)

1. I have no clue how to answer these questions so if anyone could explain how to get to the answers, that would be much appreciated.

1. What would the period of rotation of the Earth need to be if objects at the equator were to appear weightless?
radius of Earth = 6.4 X 106 m.

A 4.5 x 10-2 hours

B 1.4 hours

C 24 hours

D 160 hours

2. Two stars of mass M and 4M are at a distance d between their centres. The resultant gravitational field strength is zero along the line between their centres at a distance y from the centre of the star of mass M.
What is the value of the ratio y/d ?

A 1/2

B 1/3

C 2/3

D 3/4
2. For the first question; any object moving in a circle, regardless of the reason why it is moving in a circle, feels a resultant force that we call the centripetal force, for which there is an equation that I am sure that you know. Now, put that aside and just think what the forces acting on an object at the surface of the Earth are. Weight going towards the centre of the Earth, of course - anything else? Once you've got the resultant of the "regular" forces worked out. put it equal to the centripetal force - after all, the resultant of these forces is what we mean by the centripetal force. Finally - that "other" force from earlier - what happens to it when the Earth is spinning so fast that objects are about to be flung off? If you can see that, your equation should simplify a bit, and although there is still a little work to do, you're nearly there.

For the second question; do you know an equation for the gravitational field strength at a distance r from the centre of mass of an object of mass M? If so, work out g for both stars (in terms of y and d), and put them equal to each other (because if an object feels the same pull of gravity from both stars, but in opposite directions, it will feel a zero resultant pull). This won't enable you to find y or d, but it will let you find y/d.
3. (Original post by Pangol)
For the first question; any object moving in a circle, regardless of the reason why it is moving in a circle, feels a resultant force that we call the centripetal force, for which there is an equation that I am sure that you know. Now, put that aside and just think what the forces acting on an object at the surface of the Earth are. Weight going towards the centre of the Earth, of course - anything else? Once you've got the resultant of the "regular" forces worked out. put it equal to the centripetal force - after all, the resultant of these forces is what we mean by the centripetal force. Finally - that "other" force from earlier - what happens to it when the Earth is spinning so fast that objects are about to be flung off? If you can see that, your equation should simplify a bit, and although there is still a little work to do, you're nearly there.
The forces I can think of are:

Weight : mg
Reaction force: Idk the equation for this
Force of attraction between the object and the Earth: GMm/r2

If the objects are about to fly off, this means that there is no centripetal force so the equation will be equal to zero?

How would you find the resultant of the regular forces?

Alternatively, could you just do W = mv2/r and then substitute v for 2πr/T to work out the time period that way?
4. (Original post by jessyjellytot14)
The forces I can think of are:

Weight : mg
Reaction force: Idk the equation for this
Force of attraction between the object and the Earth: GMm/r2
Yes - but the force of attraction between the object and the Earth is the wieght of the object. You only need to use it once, so it would be better to use the mg version. You don't need an equation for the reaction force, just call it R.

(Original post by jessyjellytot14)
If the objects are about to fly off, this means that there is no centripetal force so the equation will be equal to zero?
It wouldn't be the centripetal force that would be zero, becasue that would mean that the objects are no longer travelling in a circle, and as we are considering the point where they are just about to fly off, they are still moving in a circle. So it must be one of the other forces that is changing as things go faster and faster.

(Original post by jessyjellytot14)
Alternatively, could you just do W = mv2/r and then substitute v for 2πr/T to work out the time period that way?
This is it - but where did R go? This is effectively the question I am asking above.
5. (Original post by Pangol)

This is it - but where did R go? This is effectively the question I am asking above.
I don't really know much about the reaction force, only that it balances out the weight and acts in the opposite direction. If the object is in equilibrium, the gravitational force of attraction by Earth on the object is equal to the reaction force on the object. And since the object appears to be weightless, there would be no reaction force.
Is that correct?
6. (Original post by jessyjellytot14)
I don't really know much about the reaction force, only that it balances out the weight and acts in the opposite direction. If the object is in equilibrium, the gravitational force of attraction by Earth on the object is equal to the reaction force on the object. And since the object appears to be weightless, there would be no reaction force.
Is that correct?
You are correct to say that R = mg when the object is in equilibrium. But this object is not - there is a resultant (centripetal) non-zero force acting on it. So R will not equal mg.

You are also correct to say that when an object is "weightless" (not a great phrase, unfortunately, as the object still has the same weight that it has always had as it is still at the surface of the Earth), R=0. It should now be straightforward.
7. (Original post by Pangol)
You are correct to say that R = mg when the object is in equilibrium. But this object is not - there is a resultant (centripetal) non-zero force acting on it. So R will not equal mg.

You are also correct to say that when an object is "weightless" (not a great phrase, unfortunately, as the object still has the same weight that it has always had as it is still at the surface of the Earth), R=0. It should now be straightforward.
Okay I think I understand.

The sum of the weight and the normal reaction force gives you the centripetal force.

So mg - R = m4π2r/T2

I'm not quite sure how R cancels out though.
I'm thinking it's either because R is acting in the opposite direction to the weight and the centripetal force so you just ignore it, or because the question states that the objects at the equator appear weightless so you'd assume that if there were no weight, there would be no reaction force.
But then that doesn't make sense because the object still has weight, as you said. So I'm not sure what AQA mean by 'weightless' ..

It won't let me rep you btw, so I'm not just purposefully choosing not to lol.
8. (Original post by jessyjellytot14)
I'm thinking it's either because R is acting in the opposite direction to the weight and the centripetal force so you just ignore it, or because the question states that the objects at the equator appear weightless so you'd assume that if there were no weight, there would be no reaction force.
But then that doesn't make sense because the object still has weight, as you said. So I'm not sure what AQA mean by 'weightless' ..
This is indeed confusing, but it's not AQA - this is the coloquial use of the word "weightless". Unfortunately, it doesn't mean that the object has no weight. It means that it feels as if it has no weight. Technically, it means that there is no reaction force between the object and anything that it is in contact with (well, no reaction force becasue of the pull of gravity). So put R = 0 and carry on.

This is the same thing that happens on the International Space Station, for example. Astronauts are not "weightless" because there is no gravity there. The value of g at the ISS isn't much smaller than it is on the surface of the Earth. It is becasue the space station itself and the astronauts have no reaction force between them because of gravity, as they are both accelerating at the same rate.

The same thing would happen if you were in a lift and the cable broke - both you and the lift would fall at the same rate, and it would feel to you as if you were weightless inside the lift. Well, until it hit the bottom of the lift shaft...
9. (Original post by jessyjellytot14)

But then that doesn't make sense because the object still has weight, as you said. So I'm not sure what AQA mean by 'weightless' ..
Have a look at this "weightless" picture...

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