Gravity Gradient
In standard kinematics problems involving gravity, we assume a constant gravitational field strength. Suppose instead that we have a fictitious scenario in which the gravity field strength varies as follows: g = g 0 + k y , where y (in meters) is the distance above the ground. Here, the gravity is weakest at ground level and becomes stronger with increasing altitude (whilst always pointing toward the ground).
Consider the time t f (in seconds) that it would take for an object to fall from an initial resting height y 0 to the ground. To 2 decimal places, what is the limiting value of t f as y 0 approaches infinity?
Note: The quantity g has units of m/s 2 , g 0 = 1 0 m/s 2 , and k = 1 / s 2 .
The answer is 1.57.
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4 solutions
This is a very rigorous and accurate method of calculating the time of fall. However, the time of fall for a very large initial height can be calculated directly by considering it as a simple harmonic oscillation. This is shown in the Brain's and Chirag's solution below.
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Fine, if you are only interested in answering the question and obtaining the limiting value. My solution gives so much more.
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Yes, I agree with you. Your solution gives an accurate value for any height. It is just that if you want to calculate only the limiting value then the alternative approach can be taken.
If you recognize the given differential equation as SHM, you can write the equation for y given any time and initial height. I agree that you have explicitly shown more information, but you can derive everything from SHM as well, so please don't try to downplay the method (you seem to command a fair bit of respect here)
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@Brian Moehring – By all means use SHM to obtain the explicit formula instead of solving the differential equation if you prefer. My point was merely to question why we should be content only to solve the limiting case of the problem when it is not hard to get more.
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@Mark Hennings – All right. Personally, I just see a tradeoff in utility. Your solution, while more comprehensive, must be read rather than heard. My solution has less detail, but could reasonably be explained to someone orally. I don't really see one method as better than the other, but rather complementary.
I did it the diff eq way as well. I only asked for the limiting value because I was amused by the value it took. Might be fun to come up with a harder form that forces a diff eq approach.
Although I like the spring analogy too
@Mark Hennings How did you get v 2 + ( g 0 + y ) 2 = ( g 0 + y 0 ) 2 from v d y d v = a = − g 0 − y
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Integrate the differential equation with respect to y and apply the initial condition that v = 0 when y = y 0 .
This describes a simple mass-spring system that obeys Hooke's law with spring constant k = m and equilibrium position y = − g 0 .
As we stretch the spring further, the velocity of the mass will approach infinity near the equilibrium position, so the time it takes to travel from y = 0 to y = − g 0 will approach zero, and therefore we may instead ask how long it takes for the mass-spring system to reach its equilibrium position.
Since the angular velocity is ω = k / m = 1 , the time it takes to reach the equilibrium position is ω π / 2 = 2 π .
This is a unique approach. Comparing this case with the simple harmonic motion simplifies the calculations a lot. However, you may want to note here, that for small initial heights this method will not work and the answer will depend on the initial height of drop.
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My particular answer will not apply, but I think I make it clear in the whole second paragraph that I am dealing with the limiting case.
On the other hand, the first paragraph applies for ALL initial displacements, even small ones. Therefore, if you really want to compute the time for small heights, note that the equation for the SHM in the problem can written as y + g 0 = ( y 0 + g 0 ) cos t , which we can use to solve for t when y = 0 . This will give the time for any y 0 .
Swweeeet!!!!
If we assume the motion to continue even after reaching the surface of the Earth (following the relation of acceleration given in the question), then the particle will go in simple harmonic motion.
Acceleration of the particle can be written as − a = g 0 + y .
a = − Y where, Y = g 0 + y .
Comparing this equation with equation of simple harmonic motion, a = − ω 2 y .
Here, ω is the angular frequency.
ω = 1 rad s − 1 .
In an SHM, time taken to move from one extreme to the mean position is 4 T .
Here, T is the time period equals to ω 2 π .
Therefore, the time taken to move from the extreme position to the mean position is 4 2 π .
The mean position is where the acceleration is zero, that will be at y = − g 0 .
If the object is dropped from a great height, then its speed near the Earth surface will be very large and the time to move from y = 0 to y = − g 0 = − 1 0 m will be very small. Therefore, the time to move from the infinity to the surface will be 4 2 π .
One fourth of the time period it is with angular frequency being 1 rad/s,the surface being the mean position,we can say time taken is T/4 where T is the total period,hence 2π/4=1.57,it is
I really like your thought, but this gives an answer independent of the initial height. So, are you suggesting that the answer won't depend on at which height the object is dropped?
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If you look at the Mark's solution t = 2 1 π − sin − 1 ( g 0 + y 0 g 0 This means that for small heights the answer depends on the initial height. But, If I consider your approach, it does not take care of the case of small heights. So, why your approach is not working for small heights?
It certainly does,but since the height tends to infinity the restoring force dominates/overshoots the constant force/acceleration component which is 10 m/s^2,hence even if the constant component were 25 m/s^2 the answer would pretty much remain the same
As I said the restoring force is so dominant,actually the trajectory of the particle is y=cost+10 which is the superposition of a simple harmonic motion and a constant force motion
We have the differential equation v d y d v = a = − g 0 − y so that v 2 + ( g 0 + y ) 2 = ( g 0 + y 0 ) 2 and hence d t d y = − ( g 0 + y 0 ) 2 − ( g 0 + y ) 2 Thus the time to fall to ground from height y 0 is t f = ∫ 0 y 0 ( g 0 + y 0 ) 2 − ( g 0 + y ) 2 d y = 2 1 π − sin − 1 ( g 0 + y 0 g 0 ) which tends to 2 1 π as y 0 → ∞ .