Help me calculate the value of h when I know a formula and the value of g

[keir]

Homework Statement

I have a formula and value of g but I don't know why can't I to calculate a height because everytime the result is bad.

Relevant Equations

h = (√(G * 10**-11 * M * 10**24)/g) - R

G = 6.674 - the gravitational constant
M = 5.972 - mass of the Earth
R = 6371000 - the radius of the Earth
g = 0.014 or 0.025 (I got a few results because of the different coordinates)

The result of h from my calculations is equal to about 162321138 when it should be about 400km or 400000m and I don't why.

 

[Gaussian97]

What is the original problem?

 

[TSny]

Your formula for h in terms of g looks correct. But, something's wrong with the numbers. 400 km is not very far above the surface of the earth. You can check that at 400 km, g is about 8.7 m/s².

(Actually, when I look carefully at your parentheses, they don't look correctly placed. This might be just a typo because your result of 162321138 m corresponds to using the correct formula with g = .014 m/s².)

 

[keir]

Your formula for h in terms of g looks correct. But, something's wrong with the numbers. 400 km is not very far above the surface of the earth. You can check that at 400 km, g is about 8.7 m/s².

(Actually, when I look carefully at your parentheses, they don't look correctly placed. This might be just a typo because your result of 162321138 m corresponds to using the correct formula with g = .014 m/s².)

These are the results (value of g) that are showing sensors in the computer on ISS station. When I checked the g = 8.7 the results was very close to real height of ISS station. But te results (value of g) from official page of ISS are showing very similar to my.

 

[TSny]

These are the results (value of g) that are showing sensors in the computer on ISS station. When I checked the g = 8.7 the results was very close to real height of ISS station. But te results (value of g) from official page of ISS are showing very similar to my.

Can you please give a link to the official page?

 

[keir]

Can you please give a link to the official page?

The table with results I'm talking about is on page 18
http://esamultimedia.esa.int/docs/edu/T05.3_How_to_collect_data_from_astropi.pdf

 

[TSny]

From a quick glance, this table is giving the readings of accelerometers on the ISS as multiples of g (9.8 m/s²). An accelerometer in free-fall should read zero. The accelerometers in the ISS will not quite read zero due to rotation of the station or other effects.

 

[keir]

From a quick glance, this table is giving the readings of accelerometers on the ISS as multiples of g (9.8 m/s²). An accelerometer in free-fall should read zero. The accelerometers in the ISS will not quite read zero due to rotation of the station or other effects.

So can I calculate height of ISS?
Because when I multiply 9.8 by these results in tabel and than multiply this by 10 and subtract 9.8 by previous result? and than the result is very similar to real (about 50km difference)

 

[Gaussian97]

No, you cannot, any object in free fall will have g=0 independent of the height.

 

[TSny]

So can I calculate height of ISS?

No, I don't believe so.

[In principle, the height of the ISS could be approximately determined if the accelerometers are sensitive enough to measure "tidal forces" on the ISS. See https://www.quora.com/How-much-microgravity-is-there-on-ISS-due-to-tidal-forces (particularly the answer given by Jacob Solinksky). As he shows, the tidal acceleration would be of the order of 10^-4 m/s², or 10^-5 g. But the values of acceleration given in the table in your link are at least an order of magnitude greater than this. So, the "tidal forces" appear to be too small to be measured by the accelerometers.]