Gravitational force equals centrifugal force?

[Induana]

Homework Statement

What's the reason behind fact that objects inside a spacecraft which is orbiting around Mars are weightless?

a) zero gravitational acceleration
b) Forces created by sun and earth
c) vacuum in the spacecraft and around the spacecraft
d) The result force of gravitational and centrifugal force is 0N.

Relevant Equations

F = m.a

Book says that correct answer is d) but I can't understand why. If the result of gravitational and centrifugal force is 0N then there is no force that would keep those objects inside the spacecraft orbiting around the planet. Or am I just completely wrong?

Thank you for your help.

 

[ergospherical]

In a frame rotating at the same rate as that of the spacecraft 's orbit, the spacecraft is always at rest and there is an outward centrifugal force which balances the inward gravitational force.

In a non-rotating frame, the spacecraft is performing circular motion and there is only an inward gravitational force (which is the centripetal force required to sustain this motion).

 

[haruspex]

Homework Statement:: What's the reason behind fact that objects inside a spacecraft which is orbiting around Mars are weightless?

a) zero gravitational acceleration
b) Forces created by sun and earth
c) vacuum in the spacecraft and around the spacecraft
d) The result force of gravitational and centrifugal force is 0N.
Relevant Equations:: F = m.a

Book says that correct answer is d) but I can't understand why. If the result of gravitational and centrifugal force is 0N then there is no force that would keep those objects inside the spacecraft orbiting around the planet. Or am I just completely wrong?

Thank you for your help.

You have to decide on a frame of reference.
If you use the object's frame then, by definition, it is not accelerating. Since it is an accelerating frame, we have to introduce the "fictitious" centrifugal force to make Newton’s laws work. This balances the gravitational force to produce the observed zero acceleration.
If you use an inertial frame then there is no centrifugal force, and the gravitational force results in the observed centripetal acceleration.
Options a, b and c are clearly wrong. Option d is correct if using the object's frame of reference, but not if using an inertial frane.

 

[rudransh verma]

If the result of gravitational and centrifugal force is 0N then there is no force that would keep those objects inside the spacecraft orbiting around the planet.

No. If the result of gravitational and centrifugal force in On then there is no Net force that would create weight in bodies inside the spacecraft . Weightlessness. But there are two forces due to which the body is orbiting around the planet. If one of the force increases or decreases then the orbit radius would change.
The imbalance of these two forces also causes tides on Earth on two opposite sides.

 

[haruspex]

no Net force that would create weight in bodies

How are you defining weight there?

there are two forces due to which the body is orbiting around the planet

There are? What are they?

 

[PeterDonis]

Book says

What book?

 

[PeterDonis]

there is no Net force that would create weight in bodies inside the spacecraft

Neither of the forces involved (gravitational and centrifugal--note that, as has been remarked, the latter is only present in a rotating frame) "create weight" in any situation, whether they happen to be balanced or not. (Note, for example, that in a spacecraft in an elliptical orbit, those forces will not always balance--the orbital radius changes with time--but the spacecraft is still weightless.) That's why forces like centrifugal are often called "fictitious" and why General Relativity, as opposed to Newtonian gravity, does not recognize gravity as a force and attributes it to spacetime curvature instead.

 

[rudransh verma]

How are you defining weight there?

I am defining weight as gravity, the pull of the mars.

There are? What are they?

gravity and velocity which causes centrifugal force.

 

[haruspex]

I am defining weight as gravity, the pull of the mars

Then how can the objects in orbit be weightless?

gravity and velocity which causes centrifugal force.

As I noted in post #3, you must first define your reference frame.
If you choose an inertial frame there is no centrifugal force.
If you take the orbiting object's frame of reference, or the spacecraft 's, the centrifugal force balances the gravitational force and there is no acceleration in that frame.

Weightlessness is a matter of measurement. In free fall (which includes orbiting), you cannot measure the gravitational field, but it doesn't mean there is no actual gravitational force.

 

[PeterDonis]

I am defining weight as gravity

Then you are using your own personal definition which has nothing to do with the physics definition. The physics definition is that weight is a measurable force, i.e., one that causes a nonzero reading on a scale or an accelerometer. Gravity does not. When you stand on a scale on the Earth's surface and see a nonzero weight reading, it's not the Earth's gravity that's causing the reading; it's the ground pushing up on you and the scale. If you were moving solely under gravity, you would not be standing on the Earth's surface; you would be freely falling and would be weightless.

gravity and velocity which causes centrifugal force.

It's not "velocity" that "causes" centrifugal force; it's choosing a rotating frame.

 

[jbriggs444]

Then how can the objects in orbit be weightless?

As I noted in post #3, you must first define your reference frame.
If you choose an inertial frame there is no centrifugal force.
If you take the orbiting object's frame of reference, or the spacecraft 's, the centrifugal force balances the gravitational force and there is no acceleration in that frame.

Weightlessness is a matter of measurement. In free fall (which includes orbiting), you cannot measure the gravitational field, but it doesn't mean there is no actual gravitational force.

There are at least three frames of reference that can reasonably be adopted.

1. A rotating reference frame anchored with its pivot point at the planet's center with the rotation rate selected so that the object in orbit is continually at rest. In this frame there is an outward centrifugal force and an inward gravitational force. The two are in balance. The object's gravitational weight is balanced by the outward centrifugal force.

2. An inertial reference frame in which the center of the planet is at rest. In this frame there is an inward gravitational force. This is the only force on the object. The object is subject to centripetal acceleration due to this unbalanced external force.

3. A reference frame anchored to the object. Note that this need not be a rotating reference frame. We might, for instance, adopt a coordinate system where the object is continuously at the origin and the reference frame translates in a circular pattern without rotating. In this reference frame there is a real gravitational force and a fictitious force associated with the acceleration of the coordinate system. It is not quite a centrifugal force but it does cancel the gravitational force.

When I took physics in high school, we were taught a definition of "weight" as the (opposite of the) force required to keep an object stationary in a coordinate system that one has chosen to treat as being at rest.

In this definition we carefully ignore gravity and all other inertial (aka fictitious) forces and consider only the real physical forces required to keep the object stationary. i.e. we eliminate all other physical forces, stand on an "unaccelerating" scale and see what it reads.

One's "weight" under this definition changes depending on what one chooses to regard as being unaccelerating.

 

[PeterDonis]

When I took physics in high school, we were taught a definition of "weight" as the (opposite of the) force required to keep an object stationary in a coordinate system that one has chosen to treat as being at rest.

In this definition we carefully ignore gravity and all other inertial (aka fictitious) forces and consider only the real physical forces required to keep the object stationary. i.e. we eliminate all other physical forces, stand on an "unaccelerating" scale and see what it reads.

One's "weight" under this definition changes depending on what one chooses to regard as being unaccelerating.

I'm not sure I understand the definition being given. On the one hand, you say it depends on what one chooses to regard as being "unaccelerating", which is a coordinate choice. On the other hand, you say you measure weight by standing on a scale, which means it should be a direct observable and should be independent of any choice of coordinates. Those two things don't seem to be consistent.

 

[jbriggs444]

I'm not sure I understand the definition being given. On the one hand, you say it depends on what one chooses to regard as being "unaccelerating", which is a coordinate choice. On the other hand, you say you measure weight by standing on a scale, which means it should be a direct observable and should be independent of any choice of coordinates. Those two things don't seem to be consistent.

The direct observable is only equal to your "weight" if the scale is unaccelerating.
If the scale is accelerating at a known rate, you can recover your "weight" with a calculation based on the observable together with the coordinate acceleration.

 

[PeterDonis]

The direct observable is only equal to your "weight" if the scale is unaccelerating.

In other words, you're not defining "weight" as the direct observable. It's a coordinate-dependent quantity.

If the scale is accelerating at a known rate, you can recover your "weight" with a calculation based on the observable.

So, to give a specific example, what would my "weight" be by your definition if I am standing on a scale on the Earth's surface, but I choose coordinates that are freely falling downwards? (You can leave my rest mass $m$ as an unspecified constant.)

 

[jbriggs444]

In other words, you're not defining "weight" as the direct observable. It's a coordinate-dependent quantity.So, to give a specific example, what would my "weight" be by your definition if I am standing on a scale on the Earth's surface, but I choose coordinates that are freely falling downwards? (You can leave my rest mass $m$ as an unspecified constant.)

I would be weightless under those conditions.

Obviously, it is unusual to adopt those coordinates in that situation and ask about my "weight" under the definition we are discussint.

 

[PeterDonis]

I would be weightless under those conditions.

This seems like a weird definition, then, since the scale reads nonzero and I feel weight.

 

[jbriggs444]

This seems like a weird definition, then, since the scale reads nonzero and I feel weight.

If you remove the scale, I are clearly weightless, no? So the scale plus the calculation has accurately determined my weight.

 

[PeroK]

Perhaps this thread shows that "weight," is a confusing concept in physics, and better avoided. Personally, I can't see that it adds any clarify that mass and force lack.

I would never use weight or weightless except in an everyday context.

 

[PeterDonis]

If you remove the scale, I are clearly weightless, no?

No. You're standing on the surface of the Earth. You feel weight.

 

[jbriggs444]

No. You're standing on the surface of the Earth. You feel weight.

Nope. You removed the scale. I'm not standing on anything. I'm in free fall.

 

[PeterDonis]

Obviously, it is unusual to adopt those coordinates in that situation and ask about my "weight" under the definition we are discussint.

This just makes the definition even weirder. You want to make "weight" coordinate-dependent, but you also want to rule out coordinates that would be "unusual". Why not just make "weight" the direct observable to begin with and avoid all that fuss?

 

[PeterDonis]

You removed the scale.

I didn't remove anything. I said I'm standing on a scale on the surface of the Earth, and I asked what my "weight" would be by your definition if I adopted coordinates that were freely falling. I didn't say I was freely falling, nor did I say I would be freely falling if I removed the scale. (I wouldn't--the surface of the Earth would still be there even if the scale wasn't.)

 

[PeroK]

For example. In Newtonian physics, if you do a free body diagram for an object in orbit, then there is a force $mg$ (whatever $g$ happens to be at that altitude). How is that not "weight" just as it is in a terrestrial mechanics problem?

Or, what's the weight of someone on a trampoline? Is it $mg$ or the varying reaction force? And zero when airborne?

 

[PeterDonis]

For example. In Newtonian physics, if you do a free body diagram for an object in orbit, then there is a force $mg$ (whatever $g$ happens to be at that altitude). How is that not "weight" just as it is in a terrestrial mechanics problem?

It isn't "weight" in a terrestrial mechanics problem, at least not with the definition of "weight" I'm used to. If I'm standing on a scale on the surface of the Earth, the "weight" the scale reads is not the force of the Earth's gravity pulling on me; it's the force of the Earth's surface pushing up on me. I can tell it's the latter because if I am falling freely, so that the Earth's surface isn't pushing on me but the Earth's gravity is still pulling on me, I feel zero weight.

Or, what's the weight of someone on a trampoline? Is it $mg$ or the varying reaction force? And zero when airborne?

The varying reaction force, and zero when airborne. At least, that's the definition of "weight" that I"m used to: the direct observable. Any other definition seems to me to cause more problems than it solves, for reasons I have already pointed out.