How Do You Calculate the Gravitational Field Strength on Mars?

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Homework Help Overview

The discussion revolves around calculating the gravitational field strength on the surface of Mars based on the weight of a space probe. The original poster presents a problem involving a 250-kg probe that weighs 935 N on Mars, leading to questions about the correct method for determining gravitational acceleration.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate gravitational field strength using two different methods, one involving the universal law of gravitation and the other using the relationship between weight and mass. Participants discuss the validity of these approaches and question the implications of using different values.

Discussion Status

Some participants provide guidance on which method may be more appropriate, noting that the second method directly utilizes the given weight and mass. There is an acknowledgment of the similarity in the results from both methods, prompting further exploration of the underlying assumptions.

Contextual Notes

Participants note potential confusion regarding the use of Mars' mass in calculations and the implications for understanding gravitational field strength. The discussion reflects on the importance of clarity in applying formulas and interpreting results.

ahrog
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Homework Statement


A 250-kg space probe weighs 935 N on the surface of Mars. What is the gravitational field strength on the surface of Mars?


Homework Equations


W=mg
g=Gm1/r^2
Fg=Gm1m2/r^2


The Attempt at a Solution


Okay, so at first I thought it would be:
g=(6.67x10^-11)(6.42x10^23)/(3.38x10^6)^2
g=3.75 m/s^2

BUT

Then I remembered the weight of the probe...What do I do with that?

Oh wait, brainflash.

If I went W=mg and changed it to g=W/m
g=935N/250kg
=3.74 m/s^2

...So which answer is right?
 
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In the first case you have calculated the gravitational acceleration of the probe, that is the acceleration at which a unit mass will experience if it placed a distance r from the probe in the absence of any other fields.

The latter answer is correct.

Do you now understand why?
 
Yeah, thanks! That helps! :)
 
ahrog said:
...So which answer is right?

Since both answers were the same, where was the problem?

I would go with the second solution method, since that is simpler and it makes direct use of the given information about mass and weight.
 
Redbelly98 said:
Since both answers were the same, where was the problem
How very true. I hadn't noticed that the OP had used the mass of Mars!
 

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