Initial acceleration of an object dropped above the earth

[BoldKnight399]

A object is dropped from a height of 5.29 X 10^6 m above the surface of the Earth. What is its initial acceleration. The acceleration of gravity is 9.81 m/s^2 and the radius of the Earth is 6.37 X 10^6 m. Answer in units of m/s^2.


So my problem is that I was sick today and missed the class. I can't find any problem like this in my book and my friends have no idea how to solve it either. I thought of possibly using the fact that mg=G(m1 X m2)/D^2 but that didn't help me since I am not given the mass of the object or the mass of the Earth. The only other equations I have deal with orbits, and I do not see how they would apply in this problem. I have no equation that deals with acceleration like this. (I thought about possibly using a "big four" equation, but the acceleration is not constant, so there went that possibly solving method out the window with my sanity.) If anyone has any idea how to even approach this problem my entire physics class will be in your debt.

 

[berkeman]

A object is dropped from a height of 5.29 X 10^6 m above the surface of the Earth. What is its initial acceleration. The acceleration of gravity is 9.81 m/s^2 and the radius of the Earth is 6.37 X 10^6 m. Answer in units of m/s^2.


So my problem is that I was sick today and missed the class. I can't find any problem like this in my book and my friends have no idea how to solve it either. I thought of possibly using the fact that mg=G(m1 X m2)/D^2 but that didn't help me since I am not given the mass of the object or the mass of the Earth. The only other equations I have deal with orbits, and I do not see how they would apply in this problem. I have no equation that deals with acceleration like this. (I thought about possibly using a "big four" equation, but the acceleration is not constant, so there went that possibly solving method out the window with my sanity.) If anyone has any idea how to even approach this problem my entire physics class will be in your debt.

The mass of the object doesn't matter, since F=ma. Use that equation and plug in the numbers you are given for the surface of the Earth. That will give you the mass of the Earth, which let's you use that equation in the general case, and to solve this question.

Show us your work...

 

[BoldKnight399]

hmm ok. So following the F=ma I have that:
(6.37 X 10^6)=m(9.81)
so the m(mass of the earth)=649337.4108.
Do I then plug it into the universal gravity equation thing and get:
a=G (mearth/d^2)
so a=G (mearth/(altitude dropped + radius of earth))
a=(6.673X10^-11) X (649337.4108/ 1.359x10^14)

I think I might have just made up some new rules in there...

 

[berkeman]

hmm ok. So following the F=ma I have that:
(6.37 X 10^6)=m(9.81)
so the m(mass of the earth)=649337.4108.
Do I then plug it into the universal gravity equation thing and get:
a=G (mearth/d^2)
so a=G (mearth/(altitude dropped + radius of earth))
a=(6.673X10^-11) X (649337.4108/ 1.359x10^14)

I think I might have just made up some new rules in there...

I think it's wrong from the first equation. F=ma doesn't give a force of 6.37*10^6 -- that's the radius of the Earth...

The equation you start with is:

[tex]F = m a = G \frac{m_1 m_2}{D^2}[/tex]

Plug in the values for the force at the surface of the Earth, and figure out the mass of the Earth from that. Carry units along in your equations to help you keep things straight.

 

[BoldKnight399]

gahh
ok so then I have that:
F=ma=G(m1M2)/D^2
soooo
(9.81m/s^2)=(6.673X10^-11)(M2)/(6.637X10^6)^2

right?
so then m(mass of the Earth) =6.47577X10^24kg (?)

but then how do i find the initial acceleration?
Do i plug it in again so it becomes:
a=(6.673X10^-11)X (6.47477X10^24/altitude+637X10^6)?

 

[berkeman]

gahh
ok so then I have that:
F=ma=G(m1M2)/D^2
soooo
(9.81m/s^2)=(6.673X10^-11)(M2)/(6.637X10^6)^2

right?
so then m(mass of the Earth) =6.47577X10^24kg (?)

but then how do i find the initial acceleration?
Do i plug it in again so it becomes:
a=(6.673X10^-11)X (6.47477X10^24/altitude+637X10^6)?

Much better! I used Google to check your answer for the mass of the Earth

Your last equation is pretty close -- just looks like you forgot to square the total distance term...

 

[BoldKnight399]

ok sooooo tried that...made sure to square the bottom (I got that the inital acceleration was 3.17795m/s^2) and got it wrong. I have 6 more tries and I have no idea where I went wrong or what else to do.

 

[berkeman]

ok sooooo tried that...made sure to square the bottom (I got that the inital acceleration was 3.17795m/s^2) and got it wrong. I have 6 more tries and I have no idea where I went wrong or what else to do.

Could you write out your full calculation so we can check it? See how the distance from the center of the Earth is about doubling when you add the altitude in? Since the acceleration goes as 1/D^2, you should get an answer about 1/4 of g at the Earth's surface. Looks like you aer getting more like 1/3 ish.

Also, how many numbers past the decimal point are you supposed to put in for the solution? Probably not more than a couple-three?

 

[BoldKnight399]

it has to be atleast 3 for the program to accept it, but it can be more.
ok so i did:

a=(6.673X10^-11) X [ (6.47477X10^24)/(5.29X10^6m+6.37X10^6m)^2]
a= (6.673X10^-11) X [ (6.47444X10^24)/(11660000)^2]
a=(6.673X10^-11) X [ 4.76214X10^10]
a=3.177

so what did i do wrong. Is it possibly that my mass is in the wrong units? (to be honest though, I have no idea what the units for the mass of the Earth is...)

 

[willem2]

gahh
ok so then I have that:
F=ma=G(m1M2)/D^2
soooo
(9.81m/s^2)=(6.673X10^-11)(M2)/(6.637X10^6)^2

You have an error here. That last number should be 6.37*10^6, so your mass of the
earth comes out wrong. (as google will tell you if you look at some more digits)

 

[BoldKnight399]

*cries*
ok ill try it again

 

[BoldKnight399]

omg...I am forever (rly...forever) in your debt! The answer was then 2.925m/s^2 and that was right.

 

[berkeman]

You have an error here. That last number should be 6.37*10^6, so your mass of the
earth comes out wrong. (as google will tell you if you look at some more digits)

You beat me to it (but not by much). That was a hard one to find. Thanks for the help!