All Questions
33
questions
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Is gravitational potential energy of body by $mgh$ negative? [closed]
Consider a 15kg object at 1m from earth ground level,
is P.E = 15kg * 9.8m/s^2 * 1m = 147J
or P.E = -Gm1m2/r^2 * h = -9.8 * 15kg * 1m= -147J
after browsing for a while on debate of potential energy's ...
1
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4
answers
75
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Why is work done by force $+mgh$ in the situation of throwing something up?
If there is a particle at point A(at rest) and a force moves it to point B(Above point A vertically)(final velocity = 0 at this point), the work done by gravity is $-mgh$. This I understand as the ...
- 85
1
vote
5
answers
246
views
Work done in raising an object to a height
When we raise an object to a height $h$, it is said that the potential energy of the object is increased by $mgh$. But isn't the work done by gravitational force $-mgh$?
Then that will essentially ...
- 702
0
votes
1
answer
75
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What is the significance of a reference point in calculating the potential?
The gravitational potential is given as $$U(r)=-\frac{GMm}{R}$$ where $G$ is the universal gravitational constant $M$ is the mass of the earth and $m$ is the mass of an arbitrary object and $R$ is the ...
- 335
0
votes
1
answer
47
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What happens if we define the gravitational potential zero point at a finite distance? [duplicate]
Two questions.
Can we, and if so how do we define the gravitational potential of a mass (say the sun) to be zero at a finite distance (say 1 light year)?
How does this change the gravitational force ...
- 47
4
votes
3
answers
1k
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Work done by the force of gravity
My question is very similiar to this one: Work done by gravity on falling object does not seem to equal change in mechanical energy
As I've understood it, work is only done on an object if the object ...
- 71
0
votes
1
answer
68
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Kepler's Two-Body Problem Choice of Sign
I have been going over the solution to Kepler's problem and there is a subtlety I am missing. The notes I am following (my own notes from a while ago in fact...) get to the following expression
$$
\...
1
vote
2
answers
169
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Gravitational potential energy of a two body system from infinity
In determining gravitational energy of a two body system,we define it as the negative work done by gravitational force in bringing those two bodies from infinity to a distance $r$ with respect to the ...
- 19
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votes
4
answers
124
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Is it correct to say, any object has a huge amount of potential energy?
Let's say if it is a $ 1 kg $ metal ball.
Now if we consider it together with a planet some 500 light years away (or if we consider Neptune), then there is potential energy between this metal ball ...
- 332
3
votes
3
answers
941
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What is meant by 'Gravitational Potential Energy of a System'?
'Gravitational potential energy' is defined as: 'energy an object possesses because of its position in a gravitational field'.
Consider two planets of masses $M$ and $m$ at a distance from $r$ of each ...
- 2,578
0
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2
answers
876
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Why there is a negative sign on the gravitation formula?
It all started when I was coding some simulation of the Newton's Law of Gravitation and Culomb's Law.
When I was seeking information on the internet, I found out that some people wrote this formulas ...
- 1,525
-1
votes
1
answer
35
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Potential energy in the gravitational field - Why is $r_2$ striving against infinity?
why is $r_2$ striving against infinity in the formula $π = πΊππ(\frac{1}{π_1}β\frac{1}{π_2})$, so its often simplified to $π = \frac{πΊππ}{r}$ ?
I know that in the final formula, r is the ...
1
vote
2
answers
298
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What is the height in the potential energy formulation? [duplicate]
Most examples I've seen describe the (gravitational) potential energy e.g. with an example of a ball thrown upwards, and explain how the sum of the ball's kinetic and potential energy is constant at ...
- 13
6
votes
4
answers
2k
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Why is gravity considered a negative vector in a pendulum question?
In many places and tutorials, gravity is often considered as a negative vector.
I am confused as to why is that? I though I was missing something from trigonometry but it was just negative in first ...
- 171
1
vote
1
answer
97
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Gravitational potential energy sign
Following is a small derivation just so I can explain my question. The gravitational potential energy is:
$$(*)U_g = -\frac{GMm}{r}$$
And:
$$ \Delta U =-GMm(\frac{1}{r_{final}} - \frac{1}{r_{initial}})...
- 1,038