Path of a projectile with a non-constant gravity

[ForceBoy]

The typical equation for the height of a projectile on Earth after $t$ seconds is

$h = -4.9t^{2}+vt+c$

where $v$ is the velocity of the projectile and $c$ the initial height.

This is nice and all but what happens if the height is very large? The leading coefficient of the equation above is half of the acceleration due to gravity on earth. The gravitational force between two objects of given masses is a function of the distance between the objects. If a ball is thrown into the air, the higher it gets, the less it slows down due to gravity.

I assume the distance between our projectile and the Earth is the height of the projectile. Then the equation becomes

$h = -\frac{Gm_{earth}}{2h^{2}}t^{2}+vt+c$

My problem is the graph of the equation. In terms of x and y,

$y = -\frac{C}{y^{2}} x^2+vx$

where v is a constant and C is a constant (too keep the equation general for projectiles on any planet and to avoid doing some arithmetic).

When I graph, with C and v being greater than zero, the positive part of the graph, in the first quadrant, never seems to intersect the x axis. Intuitively I believe the graph would intersect the x axis, showing there is some time for which the projectile lands. If the projectile never lands, the acceleration must be too small to keep the projectile bound. But when I make C larger, the graph just seems to grow, to expand.

So my question is whether my equation makes physical sense considering the graph suggests it doesn't (to me).

Thank you.

 

[scottdave]

One problem you have with your formula: $$ g = -\frac{Gm_{earth}}{2h^{2}}$$ it should really be $$ g = -\frac{Gm_{earth}}{2r^{2}}$$ where r is the distance between the centers of mass (from center of mass of rocket to center of mass of the Earth), not just the height above the ground.

Another thing, this formula for height comes from integrating the acceleration twice, but g is no longer a constant, which cannot just be pulled out of the integrand. This makes solving much more difficult.

How far from Earth's surface were you planning for it to go?

 

[Filip Larsen]

As general background knowledge you may also be interested to know that if you model a free falling projectile (with no air resistance) with varying gravity (using Newton's law of gravity) you effectively end up with elliptical trajectories when modeled on a round Earth (or more generally a Kepler orbit), compared to the parabolic trajectories you already know from when the projectile is modeled with constant gravity on a flat Earth.

Also, one can still model trajectories with varying gravity using flat Earth, but if speeds become significant (comparable to orbital speeds, or equivalent, that curvature of Earth becomes significant compared to the range of the trajectory) then it becomes necessary to include the (fictitious) effect of centrifugal force as well. For sub-orbital trajectories the effect of varying (Newtonian) gravity usually has much less effect on the trajectory than the effect of curving Earth, when compared to a simple parabolic trajectory on a flat Earth. In long range artillery calculations for instance, I believe you would include curvature and other effects (like the Coriolis effect, another fictitious force) long before starting to include effects of reduced gravity. In some cases, like for atmospheric models, the effect of reduced gravity is included by replacing geometric height with geopotential height.

 

[ForceBoy]

One problem you have with your formula:

g=−Gmearth2h2g=−Gmearth2h2​

g = -\frac{Gm_{earth}}{2h^{2}} it should really be

g=−Gmearth2r2g=−Gmearth2r2​

g = -\frac{Gm_{earth}}{2r^{2}} where r is the distance between the centers of mass (from center of mass of rocket to center of mass of the Earth), not just the height above the ground.

Okay. Then my distance to the center of the planet will be some $q$ and my $g$ should be

$g = -\frac{Gm_{planet}}{(h+q)^{2}}$

Another thing, this formula for height comes from integrating the acceleration twice, but g is no longer a constant, which cannot just be pulled out of the integrand. This makes solving much more difficult.

Then, would my final equation come from the following?

$h(t) =\int\int\frac{Gm_{planet}}{(h(t)+q)^{2}} dt^{2}$

And if this is so, how could I solve considering that the integrals are nested inside each other infinitely? Because $h$ is equal to the whole integral but appears inside it.

$h(t) =\int\int\frac{Gm_{planet}}{(\int\int\frac{Gm_{planet}}{\int\int\frac{Gm_{planet}}{(...+q)^{2}} dt^{2}+q)^{2}} dt^{2})+q)^{2}} dt^{2}$

As general background knowledge you may also be interested to know that if you model a free falling projectile (with no air resistance) with varying gravity (using Newton's law of gravity) you effectively end up with elliptical trajectories when modeled on a round Earth (or more generally a Kepler orbit), compared to the parabolic trajectories you already know from when the projectile is modeled with constant gravity on a flat Earth.

So, the trajectories, the parabolic ones we learn in class, are just approximations of the real elliptical trajectories?

 

[Filip Larsen]

So, the trajectories, the parabolic ones we learn in class, are just approximations of the real elliptical trajectories?

The short answer is yes, indeed. In the given context, the parabola is an approximation to the above-ground part of a sub-orbital elliptical trajectory.

However, note that both types of trajectories is a result of modeling free fall with no air resistance, so for problems involving projectiles moving through parts of the atmosphere the models quickly ends up being a fair bit more complicated, even to the point where numerical integration becomes the only feasible way of finding a trajectory solution (i.e. a closed or analytical solution is not possible or tractable). Such a situation could for instance be calculating the precise trajectory of a lifting-body vehicle reentering the atmosphere from orbital speeds, or calculating the precise helical trajectory of a spinning artillery shell (external ballistics).

 

[ForceBoy]

I will keep this in mind as I learn more. While it may not be immediately relevant, it's bound to be in the future.