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

You need to leave the denominator alone, as it is already in the form you need.P.S. I am going to assume that you really want to divide by (1.74*10^6)^2, so that the final answer would be the same as what you posted (i.e. you are not asking about why the (10^6)^2 appears in the denominator), and that you just want to know what went wrong. If I am wrong, please clarify, and I will re-post.
  • #1
terpsgirl
29
0
(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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  • #2
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.
 
  • #3
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...
 
  • #4
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.
 
  • #5
I think we're supposed to comment on:

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

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
 
  • #6
terpsgirl said:
16.2 x 1 x 10 ^-13 n/kg

What am I doing wrong??
You handled the exponents incorrectly:
[itex]10^{-11} * 10^{22} / (10^6* 10^6) = 10^{(22-11)}/10^{12} = 10^{11}*10^{-12} = 10^{-1}[/itex].
 
  • #7
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.
 

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

What is the equation for calculating the force of gravity on the moon?

The equation for calculating the force of gravity on the moon is F = G x (m1 x m2)/d^2, where G is the universal gravitational constant, m1 and m2 are the masses of the objects, and d is the distance between them.

What is the value of the universal gravitational constant (G)?

The value of the universal gravitational constant (G) is approximately 6.674 x 10^-11 Nm^2/kg^2.

How is the force of gravity on the moon different from that on Earth?

The force of gravity on the moon is significantly less than that on Earth due to the moon's smaller mass and lower surface gravity. The force of gravity on the moon is approximately 16.2 x 10^-13 N/kg, while on Earth it is approximately 9.8 N/kg.

Why is it important to solve the moon gravity problem?

Solving the moon gravity problem is important for understanding the dynamics of celestial bodies and how they interact with each other. It also has practical applications, such as in space exploration and satellite trajectory calculations.

What are some potential factors that could affect the accuracy of the calculated force of gravity on the moon?

Some potential factors that could affect the accuracy of the calculated force of gravity on the moon include variations in the moon's density and composition, as well as the presence of other large celestial bodies nearby that could also exert gravitational forces on the moon.

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