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ddddd28
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Considering a body falls a free fall and g is changing, what is the formula of the height as a function of time?
How much maths do you know? Differential equations?ddddd28 said:Considering a body falls a free fall and g is changing, what is the formula of the height as a function of time?
ddddd28 said:Unfortunately not yet, I am still in high school. Can you show me the formula and explain it briefly?
ddddd28 said:It is too advanced for me.
ddddd28 said:yes of course, calculating the velocity is not a problem at all.
ddddd28 said:As far as I understand, at the points of perigee and apogee there is no radialic speed, so, it is calculated using the circular motion, isn't it?
If your problem involves elliptical motion knowledge, as you can guess, I will not be able to solve it.
ddddd28 said:The original question was asked because of pure curiosity. Is there any ellegent proof for that? and can you express the equation by only the distance between the objects, not the surface level?
ddddd28 said:is it √GM/r ?
Others may correct me if I am wrong, but I don't think there is a direct solution for finding distance as a function of time in this situation. You can find the time to fall a given distance by single equation, but not the other way around. You get the same problem with elliptical orbits. You can directly solve for the time it takes to travel from one point of an orbit to another, but you can't do so going the other way (except for when the two points are the periapis and apoapis.)ddddd28 said:Considering a body falls a free fall and g is changing, what is the formula of the height as a function of time?
Yes, but a planet in an elliptical orbit is a free falling object.FritoTaco said:I thought you were first considering a scenario of a free falling object?
This formula only is accurate for situations where g does not change with height. The original question dealt with the situation where it does, and this is something that must be accounted for with elliptical orbits.FritoTaco said:I was in high school last year, I only have half of my notes. Is this the formula? Unless your high school is more advanced, I don't remember using perigee and apogee.
h=12gt2h=\dfrac{1}{2}gt^{2}
Janus said:Yes, but a planet in an elliptical orbit is a free falling object.
This formula only is accurate for situations where g does not change with height. The original question dealt with the situation where it does, and this is something that must be accounted for with elliptical orbits.
To calculate the height of a free-falling body, you can use the equation: h = 0.5gt^2, where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time in seconds.
The formula for calculating the time it takes for a free-falling body to reach a certain height is t = √(2h/g), where t is the time in seconds, h is the height, and g is the acceleration due to gravity (9.8 m/s^2).
No, the mass of an object does not affect its free-fall time and height. The only factors that affect the free-fall time and height are the acceleration due to gravity and the initial height of the object.
Yes, you can still use the equation h = 0.5gt^2 to calculate the height of a free-falling body with an initial velocity. However, you will need to also consider the initial velocity in your calculation. The equation would then become h = vt + 0.5gt^2, where v is the initial velocity and t is the time in seconds.
Air resistance can affect the calculation of a free-falling body's height by slowing down its descent. This means that the actual height of the object may be slightly less than the calculated height. However, for most everyday objects, the effect of air resistance is negligible and can be ignored in the calculation.