Why does bringing an object closer to Earth do negative work

[strategist]

I'm reading through Sparknotes' explanation of the equation: U = -Gm1m2/r

"Off the surface of the Earth, there’s no obvious reference point from which to measure gravitational potential energy. Conventionally, we say that an object that is an infinite distance away from the Earth has zero gravitational potential energy with respect to the Earth. Because a negative amount of work is done to bring an object closer to the Earth, gravitational potential energy is always a negative number when using this reference point."

I really don't understand why bringing an object closer to Earth would do negative work. I've been trying to wrap my head around this statement for a bit and I'd appreciate a good explanation or some guidance. From what I understand work = F*d*cos(theta) so negative work would mean pushing the object away, wouldn't it? Why would that be required to bring an object closer to Earth?

 

[Electric to be]

It is stated to be negative because no work is done against the field, rather it is the field that does work to bring it closer.

So a way to think about it is that 'we' do negative work in bringing it against the field since we allow the field to do so. However, if we were to bring the object back to the original position we would have to return this work with an equal amount of positive energy.

 

[strategist]

It is stated to be negative because no work is done against the field, rather it is the field that does work to bring it closer.

So a way to think about it is that 'we' do negative work in bringing it against the field since we allow the field to do so. However, if we were to bring the object back to the original position we would have to return this work with an equal amount of positive energy.

Sorry, but I don't really understand what you mean by the "no work is done against the field". I mean I understand that letting the object just fall requires no work but why does just "allowing" the object to fall mean we're doing negative work? In my mind, that sort of sounds like we're doing zero work but I suppose the formulas don't allow for that. Can you explain it to me in that context of the equation of W= F*d because I still don't see how we're acting opposite to the direction of motion.

I do understand that to bring it back to the original position would require an equal amount of positive energy and maybe that's all I really need to know for my exam but I think I'd rather get the whole picture because as of now I can only think of it relatively (i.e. since pushing the object away does positive work the contrary must do negative work).

 

[Electric to be]

Okay upon consideration spark notes is wrong to say that a negative amount of work is done to the mass to bring it to the earth. Gravity certainly does positive work as the kinetic energy of this mass increases as it approaches earth. By convention however, the potential energy of the gravitational system becomes more negative since the zero point is set at infinity.

 

[Drakkith]

I think part of the problem is that the explanation doesn't explicitly state who or what is doing the work. Gravity certainly doesn't do negative work, as the kinetic energy of the object is increasing and potential energy is decreasing.

Perhaps what they are trying to say is that the work done by you is negative? If you were to allow gravity to bring an object closer to Earth, and then you stop that object at some distance, the work you've done is certainly negative since the force you applied was against the direction of motion.

 

[strategist]

I think part of the problem is that the explanation doesn't explicitly state who or what is doing the work. Gravity certainly doesn't do negative work, as the kinetic energy of the object is increasing and potential energy is decreasing.

Perhaps what they are trying to say is that the work done by you is negative? If you were to allow gravity to bring an object closer to Earth, and then you stop that object at some distance, the work you've done is certainly negative since the force you applied was against the direction of motion.

I think that is what they're implying actually. It's a little weird to think about. Would it be like I'm consistently pushing upwards on a falling object but that force I'm applying is less than the force of gravity? Either way, I hope something like this isn't on the SAT Physics subject test. I don't think I'd have to calculate this kind of work especially since you can't use a calculator on the test.

 

[Drakkith]

I think that is what they're implying actually. It's a little weird to think about. Would it be like I'm consistently pushing upwards on a falling object but that force I'm applying is less than the force of gravity? Either way, I hope something like this isn't on the SAT Physics subject test. I don't think I'd have to calculate this kind of work especially since you can't use a calculator on the test.

From my experience, they usually word questions like, "find the work done by all forces", "what is the work done by gravity", or "how much work did you do to stop the object". So there usually isn't any ambiguous questions, but you never know.

 

[strategist]

From my experience, they usually word questions like, "find the work done by all forces", "what is the work done by gravity", or "how much work did you do to stop the object". So there usually isn't any ambiguous questions, but you never know.

Ah okay. So if they ask "how much work did you do to stop the object", what kind of information would you think I'd be provided? I just don't want to get caught off guard by something like this and wind up missing it on the test. The mass and radius alone wouldn't be sufficient, would they?

 

[Drakkith]

Ah okay. So if they ask "how much work did you do to stop the object", what kind of information would you think I'd be provided? I just don't want to get caught off guard by something like this and wind up missing it on the test. The mass and radius alone wouldn't be sufficient, would they?

Most likely they will want you to be able to find the difference in potential energy between two locations and perhaps use that value to determine something else.