Simulated gravity due to acceleration

I just need to have my working for the following question confirmed:

"An 81 kg astronaut is launched into space on a Saturn rocket. The upward acceleration of a Saturn rocket shortly after blastoff is 80m/s^2. When the rocket is accelerating, what is the apparent weight of the astronaut, that is, what does the astronaut experience as the weight of his body? Show your calculations(HINT: keep in mind the definition of weight and that more than acceleration is acting on the astronaut) your answer must be in newtons."

I deduced that since acceleration due to gravity is 9.8m/s, the force of the simulated gravity would be somewhere between 8 & 9 times normal due to the 80m/s^2 acceleration rate, therefore his weight would be 81kg*9.8N*8= 6350.4N.

Thank you in advance for your help.

Quote:

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Originally Posted by haflore

I just need to have my working for the following question confirmed:

"An 81 kg astronaut is launched into space on a Saturn rocket. The upward acceleration of a Saturn rocket shortly after blastoff is 80m/s^2. When the rocket is accelerating, what is the apparent weight of the astronaut, that is, what does the astronaut experience as the weight of his body? Show your calculations(HINT: keep in mind the definition of weight and that more than acceleration is acting on the astronaut) your answer must be in newtons."

I deduced that since acceleration due to gravity is 9.8m/s, the force of the simulated gravity would be somewhere between 8 & 9 times normal due to the 80m/s^2 acceleration rate, therefore his weight would be 81kg*9.8N*8= 6350.4N.

Thank you in advance for your help.

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The apparent weight of the astronaut (what a set of scales would measure) is equal to the force of the astronaut on the seat. This in turn is equal in magnitude but opposite in direction to the force of the seat on the astronaut (Newton III). Let the magnitude of this force equal R.

Net force on astronaut: Fnet = R - mg.

But Fnet = ma (Newton II).

Therefore R - mg = ma => R = ma + mg = m(a + g).

Substitute the data: R = 81(80 + 9.8) = 7273.8 Newton.

A set of bathroom scales would read (as opposed to measure) 7273.8/9.8 ~ 742 kg .......

Ok, let me just make sure I have this right...
What you're saying(in short) is that in addition to the regular force of g(9.8m/s^2) acting on the astronaut, there is the acceleration of 80m/s^2. Since he has a mass of 81Kg, we must multiply that number by the force being exerted on him(g+80m/s^2) to find his "weight".

The resulting formula is: 81kg*(g+80m/s^2)=7273.8kg*m/s^2 or 7273.8N.

His "weight" therefore is 7273.8N.

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Thank you for your help Mr. Fantastic.

Quote:

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Originally Posted by haflore

Ok, let me just make sure I have this right...
What you're saying(in short) is that in addition to the regular force of g(9.8m/s^2) acting on the astronaut, there is the acceleration of 80m/s^2. Since he has a mass of 81Kg, we must multiply that number by the force being exerted on him(g+80m/s^2) to find his "weight".

The resulting formula is: 81kg*(g+80m/s^2)=7273.8kg*m/s^2 or 7273.8N.

His "weight" therefore is 7273.8N.

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Thank you for your help Mr. Fantastic.

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I haven't checked the numbers, but this is correct in principle.

Just as an FYI: The "apparent weight" of an object is nothing more or less than the normal force on it.

-Dan

Quote:

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Originally Posted by haflore

Ok, let me just make sure I have this right...
What you're saying(in short) is that in addition to the regular force of g(9.8m/s^2) acting on the astronaut, there is the acceleration of 80m/s^2. Since he has a mass of 81Kg, we must multiply that number by the force being exerted on him(g+80m/s^2) to find his "weight".

The resulting formula is: 81kg*(g+80m/s^2)=7273.8kg*m/s^2 or 7273.8N.

His "weight" therefore is 7273.8N.

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Thank you for your help Mr. Fantastic.

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Looks OK.

The thing to remember is that weight and mass are different things. A set of scales measures weight but the scale you read is calibrated to give you mass. Nevertheless, people refer to the reading as weight, which is technically incorrect since weight is a force.

The apparent weight is as topsquark said. I'll add that if it's being measured by a spring balance, the apparent weight is the tension in the spring .....

BY the way, given the discussion so far, you might be wondering how to measure the true mass of an object when in space. There is a way but I'll let you think about it ........