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Weight of person on Earth, and on a different planet's surface?

  • Thread starter joyxx
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  • #1
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Homework Statement


A) What is the weight (in newtons) of an 80 kg person at the earths surface?
B) What is the weight of the same person on the surface of the planet with a mass 8.0 x 10^24 kg and a radius of 5.5 x 10^6 m? (G=6.67 x 10^-11 N*m^2/kg^2)
C) what is the apparent weight (scale reading) if the person stands on a newton scale inside an elevator accelerating at 0.80 m/s^2 (up)?

Homework Equations


F=ma
F=Gm1m2/r^2

The Attempt at a Solution


A) F=ma
m=80 kg
a=9.8 m/s
F=(80 kg)(9.8m/s)
=784 N

B) F=Gm1m2/r^2
G=6.67 x 10^-11 N*m^2/kg^2
m1=80 kg
m2= 8.0 x 10^24 kg
r^2=5.5 x 10^6 m
F=(6.67 x 10^-11)(80)(8.0 x 10^24)/ (5.5 x 10^6)^2
F= 4.26 x 10^16/3.025 x 10^13
=1408.26 N

C) I can't figure out the formula for this one as my lesson doesn't give out much information

This course is being done online and it is very difficult to follow and the lessons don't give out detailed descriptions on how to do the questions, and the examples aren't very helpful.
 

Answers and Replies

  • #2
berkeman
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C) I can't figure out the formula for this one as my lesson doesn't give out much information
Just add the two forces that are affecting the mass. There is a force due to the acceleration of gravity, and a force due to the acceleration of the elevator.

The forces add in this case, since the elevator is accelerating upward. Can you show that calculation? :smile:
 
  • #3
berkeman
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BTW, there is a concept called the Free Body Diagram (FBD) that we use to help us do calculations on bodies that are experiencing several forces at the same time. Have you learned about FBDs yet? :smile:
 
  • #4
Doc Al
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C) I can't figure out the formula for this one as my lesson doesn't give out much information
You'll need to apply Newton's 2nd law and solve for the normal force that acts on the person. (Just to add to berkeman's good advice.)
 
  • #5
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BTW, there is a concept called the Free Body Diagram (FBD) that we use to help us do calculations on bodies that are experiencing several forces at the same time. Have you learned about FBDs yet? :smile:
I working through my exam review, and the FBD was never in my course
 
  • #6
berkeman
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I working through my exam review, and the FBD was never in my course
That's okay, you will likely learn about those soon. For this problem, it is simple enough that you don't really need to draw a FBD. When you get to problems with several masses on tables and inclined planes all connected by ropes and pulleys, FBDs are a great way to simplify and work through those problems. :smile:
 
  • #7
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That's okay, you will likely learn about those soon. For this problem, it is simple enough that you don't really need to draw a FBD. When you get to problems with several masses on tables and inclined planes all connected by ropes and pulleys, FBDs are a great way to simplify and work through those problems. :smile:
Sorry I'm just really confused about the course, what equation should I be using?
 
  • #8
berkeman
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Sorry I'm just really confused about the course, what equation should I be using?
F = ma, just like you've been using so far. The total force on the person is the sum of the gravitational force on them, plus the force due to them being accelerated up or down by the elevator. Take a cut at the new equation so we can see it?

(and no need to be sorry, BTW)... :smile:
 
  • #9
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F = ma, just like you've been using so far. The total force on the person is the sum of the gravitational force on them, plus the force due to them being accelerated up or down by the elevator. Take a cut at the new equation so we can see it?

(and no need to be sorry, BTW)... :smile:
If I'm finding the weight the equation would be m=F/a
F=784N + 1408.26N
2192.26N
a=0.80 m/s^2
m=2192.26N/0.80 m/s^2
=2740.32
Is that right?
 
  • #10
Nathanael
Homework Helper
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If I'm finding the weight the equation would be m=F/a
F=784N + 1408.26N
2192.26N
a=0.80 m/s^2
m=2192.26N/0.80 m/s^2
=2740.32
Is that right?
784N is the persons weight on Earth right? Why should that matter?

When you stand on the ground there are basically two forces acting on you, the gravity of the planet and the normal force of the ground pushing you up. If you stand on a scale, it’ll show this normal force.

Those two forces add/subtract to give the acceleration times mass (F_net = ma). A person standing in a room is not accelerating*, so we have F_ground - F_gravity = 0. But a person in an elevator is accelerating with it, so the right side is not zero.

(*Standing in a room, you’re actually accelerating downwards according to the rotation rate of the planet, but let’s ignore that.)
 
  • #11
Doc Al
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If I'm finding the weight the equation would be m=F/a
F=784N + 1408.26N
Is the elevator in part C on earth or on the planet mentioned in part B?

Assuming that the elevator is on the earth, the answer to part B is not relevant to part C. (If the elevator is on the planet, then the answer to part A is not relevant.)
 
  • #12
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If I'm finding the weight the equation would be m=F/a
F=784N + 1408.26N
2192.26N
a=0.80 m/s^2
m=2192.26N/0.80 m/s^2
=2740.32
Is that right?
The weight is not same as mass. You are trying to find the mass of the person(which is already given). The apparent weight, measured by a scale on the elevator floor is given by the force the floor(or scale) exerts on the feet of the person.

So, to sum up, we have two forces only: The weight of the person (which is m*g) and the normal force by the floor on the person. These are oppositely directed.

The Newton's second law states that the Net force on an object is always equal to the mass of that object times the net acceleration. As the person is in the elevator, the accelerations of both the person and the elevator must be the same (otherwise the person would either hit the ceiling or sink through the floor).

And as Doc Al has said, assuming that the elevator is on Earth, the weight of the person on that other planet is not important.
 

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