What is the Power Output of Gravity?

In summary: This is also the reason that the acceleration due to gravity is greater at the poles than at the equator.Oops forgot to add those!There's N-NewtonsM-massJ-jouleS-secondsN*m-Newton massMy question is what is the power output of gravity. I'm not sure if it is just 9.81 or something else. Any guidance helps!In summary, the power output of gravity can be calculated by using the formula P=mgd/t, where P is power, m is mass, g is the acceleration due to gravity (usually taken as 9.81 m/s^2), d is distance, and t is time. However, it should be
  • #1
Kyle Wies
2
0

Homework Statement

:
It takes gravity 1.43 seconds to pull a 3.67 kg cat down from a 10m tall ledge. What is the power output of gravity? [/B]

Homework Equations


P=mgd/t
P=w/t
P=power
M=mass
G=gravity
D=distance
W=work
T=time[/B]

The Attempt at a Solution


P=(3.67)(9.81)(10)/1.43
P=251.77
 
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  • #2
Kyle Wies said:

Homework Statement

:
It takes gravity 1.43 seconds to pull a 3.67 kg cat down from a 10m tall ledge. What is the power output of gravity? [/B]

Homework Equations


P=mgd/t
P=w/t
P=power
M=mass
G=gravity
D=distance
W=work
T=time[/B]

The Attempt at a Solution


P=(3.67)(9.81)(10)/1.43
P=251.77

Do you have a question? Do your quantities ##P, M, D, W, T## have units? If so, you need to state them at some point.
 
  • #3
Oops forgot to add those!
There's
N-Newtons
M-mass
J-joule
S-seconds
N*m-Newton mass

My question is what is the power output of gravity. I'm not sure if it is just 9.81 or something else. Any guidance helps!
 
  • #4
The power output of gravity is ##P_g=\frac{dW_g}{dt}## where ##W_g## is the work done by gravity on the object. It is not "work divided by time".
 
  • #5
Kyle Wies said:
Oops forgot to add those!
There's
N-Newtons
M-mass
J-joule
S-seconds
N*m-Newton mass

My question is what is the power output of gravity. I'm not sure if it is just 9.81 or something else. Any guidance helps!

The standard value for the acceleration (##g##) of gravity (near the Earth's surface) is about 9.80665 (m/s2), which rounds to 9.81 if we want three significant figure accuracy. However, most "elementary" physics applications often take ##g = 9.8 \; m/s^2##; I, personally, use 9.81.

In fact, due to "centrifugal" effects caused by the Earth's rotation, the actual value of ##g## depends on latitude:
$$ \begin{array}{lr}
g_{\rm{poles}} &=& 9.832 \; m/s^2 \\
g_{45^o} &=& 9.806 \; m/s^2\\
g_{\rm{equator}} &=& 9.780 \; m/s^2
\end{array}$$
So, unless you are doing super-accurate ballistic or aeronautical computations, you can just go ahead and use ##g = 9.81##.

By the way: NEVER denote it by ##G## as you did; in Physics the symbol ##G## stands for the "gravitational constant", not the acceleration of gravity. People might very well subtract marks if you use ##G## instead of ##g##.
 
Last edited:
  • #6
kuruman said:
The power output of gravity is ##P_g=\frac{dW_g}{dt}## where ##W_g## is the work done by gravity on the object. It is not "work divided by time".
@Kyle Wies , in case you are confused by that post, let me clarify.
The question ought to have asked what the average power output was over the given time. You have correctly answered that.
@kuruman's point is that the power is not constant. Since the force is constant, it increases in proportion to the velocity.
 
Last edited:
  • #7
Ray Vickson said:
due to "centrifugal" effects caused by the Earth's rotation
To be accurate, the centrifugal effect changes apparent gravity, not the actual force exerted. The reduction at the equator is about 0.3%. In addition, because of the shape of the Earth, the radius is greater at the equator, reducing the actual force too - by about 0.2% compared with the poles. So the total reduction in apparent gravity is about 0.5%.
 

Related to What is the Power Output of Gravity?

What is the power output of gravity?

The power output of gravity is the amount of energy that is generated by the gravitational force between two objects. This force is dependent on the mass and distance between the objects.

How is the power output of gravity calculated?

The power output of gravity can be calculated using the formula P = F x v, where P is power, F is the gravitational force, and v is the velocity of the objects.

What factors affect the power output of gravity?

The power output of gravity is primarily affected by the mass and distance between the objects. The greater the mass and the closer the objects are, the stronger the gravitational force and therefore the higher the power output.

What is the difference between gravitational potential energy and power output of gravity?

Gravitational potential energy is the energy stored in an object due to its position in a gravitational field, while power output of gravity is the actual energy being generated by the gravitational force between objects.

Can the power output of gravity be harnessed for practical use?

Yes, the power output of gravity can be harnessed through methods such as hydroelectric power, where the gravitational force of water falling from a higher elevation is used to generate electricity.

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