As level physics question help

Samhil

4

How would you solve Q8? Thank you!

Reply 1

Absent Agent

21

Original post by Samhil

How would you solve Q8? Thank you!

Think about what would the apparent weight be if the astronaut was in a rocket accelerating upwards at a rate equal to 5g in zero gravity?

Reply 2

Samhil

OP

4

Original post by Mehrdad jafari

Think about what would the apparent weight be if the astronaut was in a rocket accelerating upwards at a rate equal to 5g in zero gravity?

Would they not appear to be weightless?

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Reply 3

Absent Agent

21

Original post by Samhil

Would they not appear to be weightless?

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It depends. They would appear to be weightless if the rocket wasn't accelerating in the absence of gravitational fields (which is fairly intuitive). But the question assumes not only that the rocket is in a gravitational field, but also that it's taking of (accelerating upwards) from the surface of the earth.

@samhill: Post edited

(edited 8 years ago)

Reply 4

Samhil

OP

4

Original post by Mehrdad jafari

It depends. They would appear to be weightless if the rocket wasn't accelerating in the absence of gravitational fields (which is fairly intuitive). But the question assumes that the rocket is taking of (accelerating upwards) from the surface of the earth.

Ah right! So as the rocket is accelerating, then the weight force of the astronaut decreases because the gravitational pull from the earth gets weaker? I'm sorry if that makes no sense as I'm still a bit new to these concepts!

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Reply 5

Absent Agent

21

Original post by Samhil

Ah right! So as the rocket is accelerating, then the weight force of the astronaut decreases because the gravitational pull from the earth gets weaker? I'm sorry if that makes no sense as I'm still a bit new to these concepts!

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You are right in thinking that the gravitational pull decreases as the astronaut is going away from the surface of the earth but the decrease in gravitational pull would be very small to be significant here and so can be neglected.

You know that a resultant force is required to accelerate a body, according to Newton's second Law F=ma, even if the body is in the absence of the gravitational fields. Therefore, in the case of an astronaut, before the take off of the rocket the weight of the astronaut is 750N and so when the rocket is taking off (accelerating upwards) there must be an external resultant force accelerating the astronaut. This external resultant force will appear to add to the weight of the astronaut if they were to stand on a scale in the rocket because bodies resist motion or change in motion.

(edited 8 years ago)

Reply 6

Samhil

OP

4

Original post by Mehrdad jafari

You are right in thinking that the gravitational pull decreases as the astronaut is going away from the surface of the earth but the decrease in gravitational pull would be very small to be significant here and so can be neglected.

You know that a resultant force is required to accelerate a body, according to Newton's second Law F=ma, even if the body is in the absence of the gravitational fields. Therefore, in the case of an astronaut, before the take off of the rocket the weight of the astronaut is 750N and so when the rocket is taking off (accelerating upwards) there must be an external resultant force accelerating the astronaut. This external resultant force will appear to add to the weight of the astronaut if they were to stand on a scale in the rocket because bodies resist motion or change in motion.

Oh right, I think I'm beginning to understand it now! So how would you work it out? Would it be something like r-750 = 5g x mass?

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Reply 7

Absent Agent

21

Original post by Samhil

Oh right, I think I'm beginning to understand it now! So how would you work it out? Would it be something like r-750 = 5g x mass?

Posted from TSR Mobile

When the rocket is not accelerating, the weight of the astronaut is W=mg. But when the rocket is accelerating upwards, the resultant force acting on the astronaut is F=ma.
Therefore the total force acting on the astronaut is:

F+W=ma+mg=m(a+g)

@samhill: Just to let you know that you don't need to find the mass of the astronaut to work out the final force. The fact that the acceleration is 5g means that the weight will be 5 times greater than the actual weight. Therefore the total weight is 6 times the actual weight.

(edited 8 years ago)

Reply 8

Samhil

OP

4

Original post by Mehrdad jafari

When the rocket is not accelerating, the weight of the astronaut is W=mg. But when the rocket is accelerating upwards, the resultant force acting on the astronaut is F=ma.
Therefore the total force acting on the astronaut is:

F+W=ma+mg=m(a+g)

@samhill: Just to let you know that you don't need to find the mass of the astronaut to work out the final force. The fact that the acceleration is 5g means that the weight will be 5 times greater than the actual weight. Therefore the total weight is 6 times the actual weight.