Question

What is the acceleration due to gravity at an altitude of 1.00 × 106 m above the earth's surface? Note: the radius of the earth is 6.38 × 106 m.

Answer 1

I have been on this problem for 30 min

Answer 2

No worries, bud. So, I'm guessing in that second line, the altitude is \[1.0\times10^{6}\]
yes?

Answer 3

yes

Answer 4

Alright. Mind me asking what formulas you think might be relevant here? Just throw out some ideas.

Answer 5

i tried to use g=GM/R^2

Answer 6

Alright, good! That's one, but you'll need one more. You'll need to get two equations involved, and use some substitution or equating in order to eliminate some things. Got an idea what the other might be?

Answer 7

I'll give you a hint: It's the one Newton is easily most famous for.

Answer 8

E=mc^2 sorry I'm lost

Answer 9

Oh, and, nevermind, actually, you had the right one. I misread, if you only have that, that's all you need.

Lol, no problem.

Answer 10

I'm going to take a little bit of a different approach and assume that you don't yet have F = GM/r^2, to make a point. Let's start out with some universal or given things, yes?

Answer 11

ok

Answer 12

use the formula g=GM/r^2

Answer 13

f is not equal to GM/r^2

Answer 14

f=Gm1m2/r^2

Answer 15

You have Newton's 2nd Law, which, when written, appears as: \[F = ma.\]This works for all objects which have a constant mass, so this should be everything you are dealing with for now. In this question, the object you are dealing with is presumably only being acted on by gravity, so you can reasonably say that \[F_{g} = mg\]Where g is the gravitational acceleration caused by Earth pulling on your object.

Next, you need to get Newton's Universal Lawof Gravitation involved. \[F = \frac {GmM}{R^{2}}\]
You can straightforward plug in Newton's 2nd Law of Motion into his Law of Universal Gravitation, cancel one of the masses, and be left with the equation you started with. That's a good starting point to get out of the way in the first place.

That's where that F = Gm/R^2 came from in the first place, so you understand. Now, onto your problem.

Answer 16

How do you think you can find the total radius, R, of the object that is some altitude above the Earth's surface? What is it?

Answer 17

6400km + given distance

Answer 18

(Answering the question for people generally isn't how people learn, but thanks...)

Answer 19

or 6.38x106

Answer 20

Rtotal= radius of earth plus altitude, so i did 1.00 10^6 + 6.38 10^6

Answer 21

thats true. i just wanted to correct the equation

Answer 22

mendicant..have you done rocket propulsion?

Answer 23

A little bit; I'm gonna try to take a shot at yours in a sec.

Answer 24

kk thx

Answer 25

So, now, you have GM/r^2, and should have values for G, M, and R, assuming you have been given (or can look up) the Gravitational Constant.\[F = \frac {GmM}{R^{2}}\]\[F_{g} = mg\]\[mg = \frac {GmM}{r^{2}} = \frac {GM}{R^{2}}\]---
\[g = \frac {GM}{R^{2}} \]

Answer 26

From there just plug and solve, yes?

Answer 27

but is R the radius + altitude

Answer 28

Physo, you misunderstand. It *is* equal to GM/R^2 in this instance. You posted Newton's Universal Law of Gravitation, which is definitely relevant, but is not the exact same ting.

Yes, R is the radius plus altitude. That's what R is. What's the problem in solving?

Answer 29

I think i might just be entering it into the calculator wrong, Thank You very much for really taking the time to help me

Answer 30

No problem! You already had the solution anyways, it seems. Let me take a shot to make sure we're getting the same thing.

Answer 31

the accelleration for all things falling or going upwards on earth is always either 9.8 m/s/s or -9.8 m/s/s

Answer 32

it actually differ by a tiny amount with altitude.

Answer 33

differs*

Answer 34

oh I never learned that!

Answer 35

\[F_g = \frac {GM}{R^{2}} = \frac {(6.67\times10^{-11})(5.97\times10^{24})}{(6.38\times10^{6}+1.00\times10^{6})^{2}} = \frac {(6.67\times10^{-11})(5.97\times10^{24})}{(7.38\times10^{6})^{2}}= \]
(Sorry, wrting it out tediously and slowly as possible)

Answer 36

so it was the notation which caused all the misunderstanding. By Fg, you mean acceleration due to gravity

Answer 37

right?

Answer 38

\[\frac {3.98\times10^{14}}{5.34\times10^{13}} = 7.31 \ \text {m/s}\]

Man, LTeX is just not playing nice tonight There we go.Did you get that?

Answer 39

And yeah, just a notation difference. *LaTeX.

Answer 40

Yes i did !!!!!!!

Answer 41

Thank You

Answer 42

No problem, man! You're welcome. You already knew how to solve it, heh, just human error.