Calculating the Value of g on Earth's Surface

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The discussion centers on calculating the value of gravitational acceleration (g) on Earth's surface, clarifying that g is an acceleration measured in m/s², not in Newtons. Participants highlight the use of two formulas: Fg = mg and Fg = (GMEm)/r², emphasizing that both can be reconciled to determine g. A misunderstanding arises when one participant attempts to calculate g using its own value, prompting clarification on logical consistency. The conversation also explains the calculator notation 'e' as shorthand for "times ten to the power of." Ultimately, the discussion concludes that both formulas can be used interchangeably to derive the same gravitational force.
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Homework Statement


What is the value of g on the surface of the Earth in Newtons?


Homework Equations


Fg=G M1M2/R^2
SumF=ma
I'm not sure which formula to use?

The Attempt at a Solution


i think its (9.80m/s^2)(5.98x10^24)= 5.8604e25 (what does the e on my calculator mean, is this how i would write the answer? something doesn't seem right.
 
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Teenytiny1991 said:

Homework Statement


What is the value of g on the surface of the Earth in Newtons?

The quantity 'g' isn't measured in Newtons. It's an acceleration. So, at least part of this question doesn't make sense.

Teenytiny1991 said:
Fg=G M1M2/R^2
SumF=ma
I'm not sure which formula to use?

You know that, close to the surface of the Earth, it's a pretty good approximation that the weight of an object (i.e. the force with which the Earth's gravity pulls on it) is given by:

Fg = mg​

where 'm' is the mass of the object, and 'g' is the acceleration due to gravity on Earth.

But you also know that, in general, the gravitational force between two masses M1 and M2 is given by Newton's Law of Gravitation:

Fg = (GM1M2)/r2

where 'r' is the distance between the two masses. This second formula applies generally i.e. it is true in all situations. In our situation, one of the two masses is the object of mass 'm', and the other mass is the Earth, whose mass we will label 'ME'. Therefore, we can re-write this formula for our situation as:

Fg = (GMEm)/r2

Now, if the second formula is always true, and the first formula (mg) is true when the object is close to the surface of the Earth, and both formulae are giving the gravitational force of attraction between the object and the Earth, then they must both be equivalent, right? Does the fact that these two formulae for calculating Fg must somehow be reconciled give you a hint as to what the defintion of 'g' must be?

Teenytiny1991 said:
i think its (9.80m/s^2)(5.98x10^24)= 5.8604e25 (what does the e on my calculator mean, is this how i would write the answer? something doesn't seem right.

Umm...you are using the value of g (9.80 m/s2) in order to try and calculate the value of g! Does anything about that strike you as logically inconsistent?

EDIT: In this context, the letter 'e' is calculator shorthand for "times ten to the power of.' For example:

4.6e5 = 4.6 x 105
 
So you are saying it doesn't matter which formula I use, because they both answer the same question. so it would be Fg= (5.98x10^24)((6.67x10^-11)=3.99x10^14
 
Teenytiny1991 said:
So you are saying it doesn't matter which formula I use, because they both answer the same question. so it would be Fg= (5.98x10^24)((6.67x10^-11)=3.99x10^14

No, I'm saying that since they both calculate the gravitational force on the object and are both true (near the surface of the Earth), then you should be able to compare them directly. After all, Fg = Fg. From that it will become clear how g must be defined if the two equations are to be consistent with each other.

Hint: both equations are of the form Fg = m*(something).
 
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