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Oct 23 2013 05:25pm
I'm not sure if this answer is correct. I have done it two ways, and came up with the same answer. However, it does not match any of the multiple choice options listed on my assignment.

Question:
With what velocity must a body be projected vertically upward from the Earth's surface, in the absence of friction, to rise to a height equal to the Earth's radius? (6400km)

I used the work energy approach, with KE = 0.5*m*v^2 and PE = m*g*h

I also used V^2 = V0^2 + 2*a*x

Both methods yielded 11200m/s

Is there something I am missing?
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Oct 23 2013 05:54pm
Write out what you have. V = ? Do not be confused, height is NOT radius. It tells you r = 6400 km. there is no f.

M = mass of earth
m = Body's mass
r = radius
1/2mv^2 - GmM/r^2 (This is the only one that relates to Earth and stuff with a radius, the other r is for centripetal force and acceleration.)


If you do the math, you get v = -Gm/R ^1/2

v =~ 7300 m/s


also in the beginning there is KE+PE, at end, PE.

This post was edited by Sefira on Oct 23 2013 06:00pm
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Oct 23 2013 06:07pm
Quote (Sefira @ 23 Oct 2013 18:54)
Write out what you have. V = ? Do not be confused, height is NOT radius. It tells you r = 6400 km. there is no f.

M = mass of earth
m = Body's mass
r = radius
1/2mv^2 - GmM/r^2 (This is the only one that relates to Earth and stuff with a radius, the other r is for centripetal force and acceleration.)


If you do the math, you get v = -Gm/R ^1/2

v =~ 7300 m/s


also in the beginning there is KE+PE, at end, PE.


What do you mean that height is not radius? The problem says that it rises to a height equal to the earths radius. I interpreted this as the object would reach a height of 6,400,000m

The answers given are:
a ) 430,000 m/s
b ) 2,600 m/s
c ) 100,000 m/s
d ) 360 m/s
e ) 7,900 m/s

This post was edited by jtd on Oct 23 2013 06:08pm
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Oct 23 2013 06:11pm
woopsies was thinking about circles, messed up there


I plugged in the values I remembered so I might've done something wrong but I am pretty sure that the formula I have up there is the correct one

For those choices, since 7,300 is what I got, since choices a/b/d are not logical, than u have either b/e

I'm going to say answer is E.

I probably did a poor explanation from above so id suggest asking professor or someone else more experienced

The two concepts we are dealing here is newton's law of gravity + conservation of energy

:ph34r:

sorry took basic physics last year, can't remember all the stuff I learned :(

This post was edited by Sefira on Oct 23 2013 06:12pm
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Oct 23 2013 06:14pm
Quote (jtd @ Oct 23 2013 07:07pm)
What do you mean that height is not radius? The problem says that it rises to a height equal to the earths radius. I interpreted this as the object would reach a height of 6,400,000m

The answers given are:
a ) 430,000 m/s
b ) 2,600 m/s
c ) 100,000 m/s
d ) 360 m/s
e ) 7,900 m/s


if you use the R= 6378100 you actually get 7,902.73 with the way they said.

This post was edited by armour on Oct 23 2013 06:15pm
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Oct 23 2013 06:32pm
Quote (Sefira @ 23 Oct 2013 18:54)
M = mass of earth
m = Body's mass
r = radius
1/2mv^2 - GmM/r^2 (This is the only one that relates to Earth and stuff with a radius, the other r is for centripetal force and acceleration.)


If you do the math, you get v = -Gm/R ^1/2


Never mind, got it. Thank you. I didn't take the square root of the whole quantity.

This post was edited by jtd on Oct 23 2013 06:42pm
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Oct 23 2013 06:52pm


I'm going to wave my hands and say that is E and I just didn't use the right constants (I mean my constants where slightly off).

This post was edited by Azrad on Oct 23 2013 06:53pm
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Oct 23 2013 06:56pm
What program is that to type out the stuff? I've seen you use it frequently, seems pretty awesome.
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Oct 23 2013 07:00pm
Its called LaTeX
It exists primarily for *nix machines, but there is a windows version
http://miktex.org/

Also there is a website where you can generate the images without an install:
http://www.codecogs.com/latex/eqneditor.php

I just used the above website, and this is the code I used in the uhh box:
Code
\\\frac{1}{2}m_2*v^2 = G*m_1*m_2*\frac{1}{r^2}\\\\
v^2 = 2*G*m_1*\frac{1}{r^2}\\\\
v^2 = 2*G*m_1*\int_{6400000}^{12800000}\frac{1}{r^2}dr\\\\
G = 6.67*10^{-11}\\\\
m_1 = 5.9*10^{24}\\\\
v\approx 7841 \frac{m}{s}
Member
Posts: 15,293
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Oct 23 2013 07:14pm
Quote (Azrad @ Oct 24 2013 01:00am)
Its called LaTeX
It exists primarily for *nix machines, but there is a windows version
http://miktex.org/

Also there is a website where you can generate the images without an install:
http://www.codecogs.com/latex/eqneditor.php

I just used the above website, and this is the code I used in the uhh box:
Code
\\\frac{1}{2}m_2*v^2 = G*m_1*m_2*\frac{1}{r^2}\\\\
v^2 = 2*G*m_1*\frac{1}{r^2}\\\\
v^2 = 2*G*m_1*\int_{6400000}^{12800000}\frac{1}{r^2}dr\\\\
G = 6.67*10^{-11}\\\\
m_1 = 5.9*10^{24}\\\\
v\approx 7841 \frac{m}{s}


Interesting. Will have to learn how to do that one day :baby:
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