Time for an object to fall to a larger one

[kaikalii]

I know that for short distances from the Earth's surface, x=-1/2gt^2+vt+x works fine for finding the time it takes for an object to fall a certain distance ignoring air resistance.
However, what if the distance is many times the Earth's radius?
The only thing I can think of to start solving this problem is f''(t)=GM/(f(t))^2, but try as I might, I cannot solve that to evaluate the time and object takes to fall, say, a distance, r. How can I accomplish this?

 

[JPaquim]

I'm assuming you're talking about a body whose mass is much smaller than the Earth's, dropped with zero velocity from an arbitrary distance from the Earth. If so, please check out this article or this one. Basically the body follows a degenerate ellipse trajectory, similar to the trajectory of the moon around the earth. The equations aren't as easy to solve as the mgh potential because they're non linear, but if you want the time in terms of initial position, you write down the energy per unit mass, which is a conserved quantity (and should be negative if you're effectively falling and not escaping the Earth): $E = E_0 = T + U = (dr/dt)^2/2 - \mu/r$, solve for dt and integrate: $$ \int_{t_0}^t dt = t - t_0 = \int_{r_0}^r \frac{dr}{\sqrt{2(E_0+\mu /r)}} = \frac{1}{\sqrt{2\mu}}\int_{r_0}^r \sqrt{\frac{r}{1-\alpha r}}dr $$ with $\alpha = -E_0/\mu > 0$, if I didn't mess up the algebra. The integral may found in integral tables or WolframAlpha, or computed numerically.

For the more general problem where you can have non-zero velocity and a body with a large mass, read into the gravitational two-body problem, or the Kepler problem if you're not interested in large masses.

 

[kaikalii]

Thank you! This is exactly what I was looking for.

 

[JPaquim]

I've added some stuff to my answer, maybe you'd like to check it out :)

 

[PJC01]

I would first clarify the question and make sure I understand the parameters and variables involved. For example, is the larger object being referred to the Earth or another celestial body? Is the distance being measured from the surface of the larger object or from its center?

Assuming the larger object is the Earth and the distance is being measured from its center, we can use the equation f''(t) = -GM/(f(t))^2, where f(t) is the distance between the falling object and the center of the Earth at time t, and GM is the product of the Earth's mass and the gravitational constant. This equation is known as the universal law of gravitation and it applies to all objects in the universe.

To solve this equation, we can use numerical methods such as Euler's method or Runge-Kutta method to approximate the solution. These methods involve dividing the time interval into smaller intervals and calculating the distance at each time step. This approach can give us a good estimate of the time it takes for an object to fall to a larger object, even for distances much larger than the Earth's radius.

Another approach would be to use the concept of escape velocity. If the distance between the two objects is much larger than the Earth's radius, we can assume that the falling object has enough kinetic energy to escape the gravitational pull of the larger object. In this case, we can use the equation for escape velocity, v = √(2GM/r), where v is the escape velocity, G is the gravitational constant, M is the mass of the larger object, and r is the distance between the two objects. We can then use this velocity to calculate the time it would take for the object to reach the larger object's surface.

In summary, the equation f''(t) = -GM/(f(t))^2 can be used to calculate the time it takes for an object to fall to a larger object, even for distances much larger than the Earth's radius. However, numerical methods or the concept of escape velocity may be needed to solve this equation for specific distances.