Solving Moon Gravity Problem: 16.2 x 10^-13 N/kg

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The discussion focuses on solving the Moon's gravity problem using the formula for gravitational acceleration. The calculation involves the gravitational constant, the mass of the Moon, and the distance from its center, leading to a simplified expression of 16.2 x 10^-13 N/kg. Participants highlight issues with exponent handling and suggest using a calculator for accuracy. The main concern is ensuring proper simplification and maintaining correct units throughout the computation. The conversation emphasizes the importance of clear notation and accurate mathematical operations to avoid confusion.
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(6.67 x 10 ^-11 N m^2/kg^2) (7.36 x 10^22 kg)
___________________________________________
(1.74 x 10^6m) ^2

simplify:

6.67 x 7.36
___________ x 10^-11=22-6(2) x Nm^2/kg^2 * kg
1.74 ^2 ____________ _____________
(10^6)^2 m^2

16.2 x 1 x 10 ^-13 n/kg



What am I doing wrong??
 
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Have a look at what you wrote, it's impossible to read. Moreover, I have no idea what you're doing, and why. What is the actual question you're trying to answer? Is this simply an error in your computation? If so, I suggest you use a calculator.
 
This problem was given on a previous test...the problem is written exactly as seen on the actual exam:

Moons gravity is (6.67 x 10 ^-11 N m^2/kg^2) (7.36 x 10^22 kg)
___________________________________________
(1.74 x 10^6m) ^2


I was to simplify and give units...I ended up w/ an incorrect answer and I did all the steps...
 
1.62 N/kg

Look at the dimensions, you'll see that it ends up being N/kg. Use a calculator, and you'll get a number like:

1.6214559386973180076628352490421

Only keep 3 significant digits and you'll get what I got. Were you allowed to use a calculator? If so, I don't see what was so hard. If not, the question is ridiculous.
 
I think we're supposed to comment on:

\frac{(6.67 * 10^{-11})*(7.36*10^{22})}{(1.74*10^{6})^2}

and its simplification to, hopefully, whatever the result of (6.67*7.36)/(1.74)^2 is times 10^{-1}
you appear to have used the 10^{6} in the denominator twice.
note you can't assume that what you type in the white box will come out the same in the finished article; fixed width fonts anyone? plus all spaces are treated equally, I think. click on the pretty tex to see how to do it
 
terpsgirl said:
16.2 x 1 x 10 ^-13 n/kg

What am I doing wrong??
You handled the exponents incorrectly:
10^{-11} * 10^{22} / (10^6* 10^6) = 10^{(22-11)}/10^{12} = 10^{11}*10^{-12} = 10^{-1}.
 
Specifically, you started off good, then did too much of a good thing.

You started off with 10^ [-11+22-6(2)], which is the right way to handle the exponents ( I think, assuming you meant + instead of =).

Then you subtracted the exponents in the denominator again.
 
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