# How can a rocket which is angled at 30° in relation to the Earth be able to fly?

Someone told me that SpaceX's Falcon Heavy would fall out of the sky because of the angle the rocket was at (at 125 seconds after launch). The angle was 30 degrees,the acceleration was ~25 meters per second, the velocity was ~4200 km per hour.

The equations they used are as follows:

Resultant Acceleration up $= \tan(30°)/1.01=0.583g$. That means resultant acceleration downward $1g-0.583g=0.41g$ which will head the rocket straight for the ground.

I'm not familiar with these equations and how they work.

The numbers for the angle,accerlation,and velocity were taken from the telemetry data, found on this Reddit thread.

• By this logic, the Apollo rockets could never reached the Moon because their acceleration at launch was $2.2\,\text{m/s}^2$. The $F$ in $F=ma$ is the net force. To have an upward acceleration of $2.2\,\text{m/s}^2$, the upward force by the rocket necessary had to have exceeded downward force by gravitation. The same goes for the Falcon Heavy, which was well into its gravity turn at T+125 seconds. – David Hammen Feb 19 '18 at 10:00

## 2 Answers

For an acceleration of 25 m/s$^2$, and an angle of 30 degrees (since you didn't specify whether it was 30 degrees from the vertical or from the horizontal, I'm assuming the worst-case scenario of 30 degrees from the horizontal), the upward acceleration is $25\sin (30°)=12.5$ m/s$^2$, still substantially greater than $g=9.8$ m/s$^2$.

But let's assume your friend didn't make an arithmetic mistake for a moment. Why do they assume that a net downward acceleration means a rocket is headed straight for the ground? Anything in orbit always has a net downward acceleration. In fact, objects in low-Earth orbit have a net downward acceleration of $1g$! Yet they still manage to stay above the surface of the Earth pretty much indefinitely (and the Earth manages to stay out of the Sun, etc.).

This is because there are two quantities that are important when determining the flight time of an object: its acceleration and its velocity. After all, acceleration only gives you the rate of increase or decrease of the velocity, so if an object has a bunch of initial velocity, it's not coming down anytime soon (maybe ever, if it's fast enough). Objects in orbit have a high enough horizontal velocity that the acceleration in the vertical direction never actually manages to bring them down to the surface. It merely changes the direction that their velocity vector is facing.

• Oh yes sorry,I think it would be called from the horizontal(as it is a rocket trying to achieve orbit.) – Austin Brady Feb 19 '18 at 7:10
• @AustinBrady Doesn't really change the answer, in any case. Taking it from the vertical would make the upward acceleration even larger. – probably_someone Feb 19 '18 at 7:11

Your friend's claim isn't quite correct for the time slot he gives, but overall the argument that the rocket would "fall out of the sky" because it has a negative vertical acceleration is just plain wrong.

The initial stages of the rocket launch are focused on obtaining a very large vertical speed (at T+125s, some 800m/s), but the rocket's final goal (at least before its eventual trans-Mars injection burn) is a circular orbit at a constant distance from the ground and therefore at zero vertical velocity. That means that vertical deceleration is inevitable and required in the later stages of any rocket launch to orbit; this deceleration is normally enacted by the Earth's gravity sapping the initial vertical speed as the rocket goes up. (Much like a ball that's thrown up on the ascending part of its arc, the fact that it's accelerating downwards doesn't mean it's moving downwards, it just means it's losing vertical speed.)

This process can be seen very clearly on the telemetry data you've linked to: if you go to the velocity vs time link here, you'll see the graph I reproduce below, where the green line indicates the vertical component of the velocity. Note in particular that from T+180s that green line is on a monotonous descent: this is no more and no less than vertical deceleration. OK, so, having established that the rocket is indeed accelerating downwards, then why doesn't it fall out of the sky? The answer is in that sideways tilt of the rocket: you need a lot of vertical speed to get to orbital height, but what you really need is lots of horizontal speed to actually stay in that orbit.

Ultimately, the answer boils down to the same question for Newton's cannonball, shown below, which does not have a rocket engine and which therefore experiences the full brunt of the gravitational acceleration towards the center of the Earth, but which is going so fast that the ground curves away under it as it tries to "fall out of the sky". The same thing is happening with the rocket, for which the second stage's cutoff acts like the end of the gun barrel for Newton's cannonball. This is carefully timed with the upwards end of the vertical arc, i.e. when the downwards acceleration of gravity finally gets rid of the last bit of upwards vertical speed, the rocket has achieved so much horizontal velocity that it is no longer useful to think of gravity as a vertical force and you need to think of it as the radial centripetal force that keeps the rocket from flying off into infinity.

For more details, a good resource is the xkcd What if? piece on Orbital Speed.