1. Oct 11, 2012

### christina007

I am absolutely new to this. I just started and my professor is popping questions out and I know NOTHING about anything yet!! I think I was placed in the wrong class because everything has completely went over my head! I don't even know where to begin to even try to start this... my professor won't even explain to me how to do it. It's very discouraging. We aren't allowed to ask questions during lecture and afterwards he has another class to get to and won't meet with me. I just don't know what to do. :( I would use the default given to post questions but I still don't even know what to attempt to solve or anything relevant. I am 100% sure I need to get out of this class and into the right one ASAP. Can anyone please try to help me with this question and explain how you got it so I don't feel like a complete idiot.

The answer should be 7.3, I just don't know how to figure it out. :(

6.67x10E-11 N-m^2/kg^2)x(5.98x10E24kg)
------------------------------------------=
(1.01x10E6m + 6.38x10E6m)

***OR***

6.67x10^-11 N-m^2/kg^2)x(5.98x10^24kg)
-------------------------------------------=
(1.01x10^6m + 6.38x10^6m)

And it is for acceleration of a satellite above Earth?

Monday I am going to the school and getting them to check and look at all of my things and see if I can get in a complete beginner Physics class. :(((((((

2. Oct 11, 2012

### Sourabh N

Do you mean you cannot figure out how to arrive that expression* or how to get to 7.3 from that expression*?

*that expression = $\frac{(6.67*10^{-11} Nm^2/kg^2)(5.98*10^{24}kg)}{(1.01*10^6m + 6.38*10^6m)}$

Last edited: Oct 11, 2012
3. Oct 11, 2012

### christina007

Thanks! I honestly don't know, but I'm going to assume and go with how to get 7.3
I am a lost cause with Physics already! :(

This is how he worded it and left it. Unable to ask questions or anything. His logic is "competent enough to make it into my class, competent enough to figure things out"
(copying the way you wrote it, ty)

$\frac{(6.67*10^{-11} Nm^2/kg^2)(5.98*10^24kg)}{(1.01*10^6m + 6.38*10^6m)}$ = 7.3 How?

Last edited: Oct 11, 2012
4. Oct 11, 2012

### christina007

sorry, double posted :x

5. Oct 11, 2012

### Sourabh N

I am sorry about your situation with the professor, but as far as this question is concerned, it's just arithmetic. I'll give you a crash course:

1. Bring all the units out.
That means you separate numbers from letters. This expression would look like:

$\frac{6.67x10^{-11} N-m^2/kg^2)x(5.98x10^{24}kg)}{(1.01x10^6m + 6.38x10^6m)} \rightarrow \frac{6.67x10^{-11} )x(5.98x10^{24})}{(1.01x10^6 + 6.38x10^6)} \frac{\frac{N m^2}{kg^2} kg}{m}$

2. Do the arithmetic, solve the units.

Try this and see where you get stuck.

Also, regarding the lack of help, try to avail the services of physics and math tutors. Most Physics and Math departments (at least in US) have free tutoring available.

6. Oct 11, 2012

### CWatters

Can't get "7.3" from that expression. Must be something missing.

Suggest you have a chat with another student and ask to see their notes?

Last edited: Oct 11, 2012
7. Oct 11, 2012

### Sourabh N

Thanks CWatters. Never did the calculation, glad you point that out.

8. Oct 11, 2012

### CWatters

Got it....

The equation appears to be something like...

=(Gravitational constant G * mass of the earth)/(height of sat? + radius of the earth)

In which case I believe the equation should actually be..

=(Gravitational constant G * mass of the earth)/(height of sat? + radius of the earth)2

The normal form is..

=Gm/r2

If you do that then you do get the answer 7.3

9. Oct 11, 2012

### Simon Bridge

Aside:
Lets take a break from our regular programming for this word from our sponsor...
Tired eyes? Math formatting too small? Try regular tex mode like this:

$$\frac{(6.67\times 10^{-11}) \text{Nm}^2\text{kg}^{-2}\cdot (5.98\times 10^{24})\text{kg}}{(1.01\times 10^6)\text{m} + (6.38\times10^6)\text{m}}$$
- using tex instead of itex lets you write the equation out full-sized.
- using \times instead of * gets you $\times$ instead of $*$ - the latter being the sign for a convolution.
- using \text{units} puts units upright so they don't get confused with math variables and constants (though it can get a bit annoying to write.)

Whether you use the brackets just around the number or around the units as well is more a matter if taste, as is the \cdot between the terms in the numerator ;)

It's easy! Just click the "quote" button at the bottom of this post to see how I did that :D

Why the "?" - don't you know? Anyway ... acceleration, in SI units, should be $\text{ms}^{-2}$ or $\text{Nkg}^{-1}$ because $F=ma$.
Your units come out to $\text{Nmkg}^{-1}$ and squaring the denominator fixes it.