Finding Altitude When Given g = g0(Re/Re+A)^2

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In summary, the conversation discusses the calculation of altitude based on the acceleration due to gravity, g0, the altitude A, and the radius of the earth, Re. Using the given equation and known values, the altitude can be determined. The process involves canceling out units, taking the square root, and converting units to find the final answer.
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Oannes
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Homework Statement


[/B]
g = g0(Re/(Re + A))2

g is the acceleration due to gravity. g0 is the acceleration of gravity at the surface of the earth, A is altitude, and Re is the radius, approximately 6380 km. Assume g0 = 9.8 meters per second squared. If the value of g is 9 meters per second squared, what is the Altitude in units of miles?

Homework Equations


[/B]
g = g0(Re/(Re + A))2

The Attempt at a Solution



First, I plugged what I knew into the given equation:

9m/sec2 = 9.8m/sec2(6380km/(6380km +A))2

Second, I divided m/sec from the right:

(9m/sec2)/(9.8m/sec2) = 0.9583
0.9583 = (6380/(6380 + A))2

Third, I took the square root of both sides:

0.9583 = 6380km/(6380km + A)

Fourth, I multiplied both sides by the denominator:

(6380km + A)(.9583) = (6380km/(6380km + A))(6380km + A)
6113.954km + .9583A = 6380km

Finally, I subtracted, divided, and converted units:

.9583A = 266.046km
A = 277.6229km

277.6229km *(.621 mi/ 1km) = 172.404 miles

My Questions: So while typing this problem up I started to understood more of what it is I needed to do and it became much easier. However, I still have some questions about converting the units and so forth.First, during the second step where I divided m/sec on the right side from the left side I canceled the units out during division as if they were a variable, I am pretty sure this is legal but I want to make sure that I am right in this assumption, otherwise I would have to convert m/sec^2 to km and I have no idea where to start on that one. Second, is taking the square root of both sides to get rid of the exponent a viable option? Once again I feel like this is true, but I am very paranoid and cautious when it comes to these things. Thanks in advance for any help given
 
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  • #2
Oannes said:
First, during the second step where I divided m/sec on the right side from the left side I canceled the units out during division as if they were a variable, I am pretty sure this is legal but I want to make sure that I am right in this assumption,
Yes, that is what you need to do. It gives you an answer that is a dimensionless❇ ratio
Second, is taking the square root of both sides to get rid of the exponent a viable option?
Yes, take the square-root of everything under a single big square-root sign, for each side.

❇ this means it has no units
 

1. What is the equation for finding altitude when given g?

The equation for finding altitude when given g is: g = g0(Re/Re+A)^2, where g0 is the acceleration due to gravity at the Earth's surface, Re is the radius of the Earth, and A is the altitude above the Earth's surface.

2. How is this equation derived?

This equation is derived from the law of universal gravitation, which states that the force of gravity between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. By setting this force equal to the centripetal force required to keep an object in circular motion, we can solve for the acceleration due to gravity at a certain altitude.

3. What units are used in this equation?

The units used in this equation will depend on the units used for the constants g0, Re, and A. Typically, g0 is measured in meters per second squared (m/s^2), Re is measured in meters (m), and A is measured in meters (m). Therefore, the resulting units for g will be in meters per second squared (m/s^2).

4. Why is this equation important in scientific research?

This equation is important in scientific research because it allows for the calculation of altitude based on the acceleration due to gravity. This can be useful in a variety of fields, such as aerospace engineering, geology, and atmospheric science. It also helps to understand the effects of gravity at different altitudes, which is important for space exploration and satellite orbits.

5. Are there any limitations to this equation?

Yes, there are some limitations to this equation. It assumes a spherical Earth and does not take into account variations in the Earth's gravity field due to factors such as topography and density. It also does not account for other forces that may affect an object's motion, such as air resistance. Additionally, this equation is only accurate for altitudes close to the Earth's surface and may not be applicable at extreme altitudes or in non-Earth environments.

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