I was explaining to my 8 year old daughter that objects in free fall follow an elliptical trajectory instead of the commonly believed parabolic one (source: https://www.forbes.com/sites/startswithabang/2020/03/12/we-all-learned-physics-biggest-myth-that-projectiles-make-a-parabola/). I told her that only on a flat Earth would an object on free fall follow a parabolic trajectory. Then she asked me what shape the Earth would have to be for an object in free fall to follow a straight line trajectory. Is it even possible?
[Physics] What shape the Earth would have to be for an object in free fall to follow a straight line trajectory
free fallnewtonian-gravityprojectile
Related Solutions
For the record, here's a worked solution:
If $r$ is the distance between the two point masses $m_1$ and $m_2$--which start at rest--then as both accelerate towards each other, where
$$ \frac{d^2r}{dt^2} = -\frac{Gm}{r^2} \ \ \ , \ \ \ \ \ m = m_1 + m_2$$
The first step in solving this equation is the least obvious: Multiply both sides by $\displaystyle \frac{dr}{dt}$ and integrate from time $0 \rightarrow t$. Writing $v = dr/dt$, the limits in velocity are $0 \rightarrow v = v(t)$, we have on the left hand side
$\displaystyle \int_0^t \frac{d^2r}{dt^2} \frac{dr}{dt} dt \ = \ \int_0^t \frac{dv}{dt} . v \ dt \ = \ \int_0^t v . \frac{dv}{dt}\ dt \ = \ \int_0^v v \ dv \ = \ \frac{1}{2}v^2 \ = \ \frac{1}{2} \left( \frac{dr}{dt} \right)^2$.
On the right hand side, writing $R$ for the starting distance, we integrate $R \rightarrow r = r(t)$,
$\displaystyle \int_0^t - \frac{Gm}{r^2} \frac{dr}{dt} dt \ = \ -Gm \int_R^r \frac{dr}{r^2} \ = \ Gm \left( \frac{1}{r} - \frac{1}{R} \right)$.
Putting these two expressions together and choosing the negative square root as $v = dr/dt$ is negative, growing in magnitude, we have
$$ \frac{dr}{dt} \ = \ - \sqrt{2Gm} \sqrt{ \frac{1}{r} - \frac{1}{R} }.$$
We are now in familiar territory as this equation is separable. Integrate once more, with limits $t = 0 \rightarrow T$ and $r = R \rightarrow 0$, and we arrive at the collision time of
$$ T = \frac{\pi}{2} \sqrt{\frac{R^3}{2Gm}}.$$
The last step uses the integral
$$ \int \frac{\sqrt{r}}{\sqrt{R-r}} dr \ = \ -\sqrt{r}\sqrt{R-r} + R \arcsin \sqrt{\frac{r}{R}} \ \ \ \ (\ +C\ ) \ .$$
This integral is sometimes calculated and written with $\arctan$, but the form here with $\arcsin$ is slightly easier for our purposes as it avoids dealing with the limit discussed above.
When I arrived at the expression for collision time $T$, I was suspicious. I wrote a simple numerical simulation, and yep--it holds up.
I can't say I have ever seen this formula for collision time anywhere. If I ever have the privilege of teaching ODEs again, this would make a great problem.
Now that doesn't apply on the orbit of the Earth. The gravitational force can be thought of as constant since the distance fron Earth to Sun can be thought of as constant too
You are correct that the strength or magnitude of the sun's gravitational field is very similar over the length of the earth's orbit, but the direction is not. In a uniform gravitational field, the direction would be the same everywhere.
Over the path of the earth's orbit, the sun's gravitational field points in different directions. This significant difference from a uniform field means that the earth's orbit is quite far from a parabola.
Best Answer
Yes, it is possible, under very strange (effectively purely hypothetical) circumstances.
Suppose that the shape of the earth was a uniform-density hollow spherical shell and suppose that instead of living on the outside of the earth, we lived on the inside of the shell. In this case the trajectory of a projectile will be a straight line.
The reason there is a straight line trajectory in this case is because in this case there is no gravitational force on the projectile (since there is no mass within the inner part of the sphere and since the force from the shell conveniently happens to exactly cancel everywhere within the shell).