# Falling to Earth from Orbit [closed]

Suppose we ascend to a height $$h$$ away from the surface of the earth and drop an object. High school physics tells us that it will accelerate at a speed of $$9.8 \frac ms$$ and will hit the ground in $$\sqrt{\frac h{4.9}}$$ seconds. However, if we ascend high enough, the gravitational force from the earth is less, and so the acceleration is less, and it will take longer to hit the ground. What is the true equation for how far the object has fallen and how long it will take to reach the ground?

Here's what I have so far: for generality, let the earth or other planet have a radius of $$r$$ and surface acceleration due to gravity of $$G$$.

We can derive from the variables above that the force acting on the object at height $$h$$ is $$mG\left(\frac r h\right)^2$$. Thus it will accelerate at $$G\left(\frac r h\right)^2 = Gr^2 \frac 1 {h^2}\frac m{s^2}$$. I want to integrate this with respect to time, and yet my variable is in meters. I need some related rate between the two, but that related rate would be how many seconds it takes to fall some number of meters, which is exactly what I am looking for to begin with. How can I proceed?

• Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions over those just asking for a specific computation. Commented Jan 3, 2022 at 20:13
• Does this answer your question? The Time That 2 Masses Will Collide Due To Newtonian Gravity Commented Jan 3, 2022 at 20:42
• Re "...accelerate at a speed of 9.8 m/s": That is confusing. E.g., that is not the unit for acceleration. Speed (velocity?) and acceleration are two different things - do you mean integral or derivative/rate of change? Can you clarify by rephrasing your question? (But without "Edit:", "Update:", or similar - the question should appear as if it was written right now). Commented Jan 4, 2022 at 12:46
• I have a spreadsheet that gives a numeric answer for this problem. You put in values for the initial conditions. It works with a series of steps going down. Conservation of energy gives the velocity at each point. The force equation gives the acceleration. With these, the time for each short segment of the drop is calculated, and a cumulative sum gives the total time. Commented Jan 4, 2022 at 20:02
• If your coming down from an orbit, you lose orbital speed. That puts you into an elliptical orbit which intersects the surface. Then you use friction with the air to lose the rest of your speed. Commented Jan 4, 2022 at 20:14

Since you're a math major, the way to think about this is that Newton's Second Law is an ODE. If $$h$$ is the distance between the object and the center of the Earth, the force on the object is (as you have shown) $$F = - m g \frac{r^2}{h^2}$$ where $$r$$ is the radius of the Earth. (Note the minus sign, since the force is attractive. Also, $$g$$ is more standard for the gravitational acceleration; $$G$$ is the gravitational constant, which is different.) And since $$F = m a$$, we have $$m \frac{d^2h}{dt^2} = - m g \frac{r^2}{h^2} \quad \Rightarrow \frac{d^2h}{dt^2} = - \frac{g r^2}{h^2}$$
Now you can break out your skills from your ODEs class to solve this ODE for $$h(t)$$, subject to the initial conditions of $$h(0) = h_0$$ (some initial position) and $$h'(0) = 0$$ (the vertical velocity of the particle is zero.)
I will warn you that (IIRC) there is not a closed-form expression for $$h(t)$$ in terms of elementary functions that solves this equation. But there is an (ugly) expression for $$t(h)$$—and since you want to find the amount of time for a given $$h_0$$, that's probably more useful to you anyhow.
• @WHG: I was using $h$ throughout to denote the distance between the object and the center of the Earth, in order to remain consistent with the OP's work. Commented Jan 4, 2022 at 0:36