Meteorites leave strewn-fields before impacting, SpaceX rockets come in at an angle and stand up straight at the last second, the Soyuz capsule doesn't fall straight. I assume this has to do with the angle needed to penetrate the earth's atmosphere.

But then I was thinking about how SpaceX lands it's rockets and how movie aliens/spacecraft tend to fall straight down from the sky. And they fall down straight from the sky because there wasn't the example of SpaceX landing spacecraft before.

I also assume that SpaceX could do a more vertical landing but that would require more fuel after reentry.

I always see the meteor that destroyed the dinosaurs depicted this way - falling straight down - but is that an artistic choice/void of information or scientifically accurate?

So what are the requirements for something to fall straight down to earth?

  • $\begingroup$ That it not explode. This all depends on how fast it's moving during reentry. If solar-system type speeds, an Earth Shattering Kaboom will surely ensue! $\endgroup$
    – uhoh
    Feb 9 '18 at 8:30
  • 2
    $\begingroup$ "And they fall down straight from the sky because there wasn't the example of SpaceX landing spacecraft before" - Didn't need to wait for SpaceX to figure out that things don't fall vertically from the sky. Artists making low-fidelity space movies tend to not bother working out the physics behind what they're depicting. $\endgroup$
    – Hobbes
    Feb 9 '18 at 11:51
  • $\begingroup$ @Hobbes but they do base things on things they have observed. why are phasers usually shaped like guns instead of something else? $\endgroup$ Feb 9 '18 at 19:23
  • $\begingroup$ SpaceX could stick a barge more or less directly underneath where a booster/lower stage decouples and then drop the rocket straight down and it would be more fuel-efficient than RTLS to the LZ. But returning to LZ optimizes for ease of re-use, while putting the barge further downrange maximize how much it can throw and still be reusable. Dropping it straight down doesn't solve any problems. Kicking it back to the LZ so hard that it would drop perfectly vertical would just waste fuel and reduce how much it could throw and still get back. $\endgroup$
    – lamont
    Feb 9 '18 at 19:27
  • $\begingroup$ Long before there were rockets, we could observe meteorites falling - and they all fall in a long arc, not vertically. When a space movie shows a ship falling vertically, the author either hasn't spent any thought on realism, or he's postulated antigravity propulsion. $\endgroup$
    – Hobbes
    Feb 9 '18 at 19:46

Sure it's possible.

There is two ways for this to be achieved:

  • For an earth satellite; cancel all forward velocity.

The satellite will then start being attracted by earth gravity and fall straight down (minus moon's orbital perturbation, ...)

  • For an object coming from outside the earth gravity well, to move faster or slower, in a similar axis as the earth at the time of impact.

The earth will "catch up" or the asteroid will catch up with earth.

Due to the gravity exerced by the earth, sun, ... This also won't be exactly a straight line, but this might looks like one.

Most meteors arrive at an angle, so the straight up meteor fall is probably not realistic even tho it's not entirely impossible

I recommend you watching this video, makes gravity much more intuitive:

  • $\begingroup$ Really nice demo in the video $\endgroup$ Feb 9 '18 at 13:40
  • 2
    $\begingroup$ "The satellite will then start being attracted by earth gravity" - s/start/continue/? $\endgroup$
    – npostavs
    Feb 9 '18 at 13:51
  • 2
    $\begingroup$ Canceling a satellite's forward velocity still wouldn't result in it falling straight down, right? The Eötvös effect should still cause an eastward deflection. $\endgroup$
    – Nobody
    Feb 9 '18 at 14:27
  • 2
    $\begingroup$ "cancel all forward velocity." and do it quickly, since as soon as you slow down just a little, you'll start arcing down. $\endgroup$
    – RonJohn
    Feb 9 '18 at 15:42
  • $\begingroup$ XKCD reference $\endgroup$
    – JBentley
    Feb 9 '18 at 19:27

Obligatory XKCD

Forgive me for assuming this, but it seems like you don't quite understand what an orbit is. As XKCD Puts it "Space is not up, space is sideways very, very fast."

To put it another way, if I hold a ball in my hand and drop it, it will go straight down. If I flick the ball a little bit, it'll follow a very steep arc - still mostly down. If I throw the ball as hard as I can, it will take a very long and shallow arc, almost a straight line at first. If I put that ball on a rocket and send it on arc over the horizon, the balls trajectory will match the curvature of the earth, and it will never come down - the ball has achieved an orbit.

Here's the problem. The sideways speed required to get into low earth orbit is ~4.9 miles per second. That is very, very fast, and requires a comparably large amount of fuel to get up to that speed. Also, given that you can't just stomp the brakes in space, you need to bring enough fuel to slow back down. If I want to come straight down, I have to cancel all of that speed out, doubling the delta-V requirement and exponentially increasing the fuel mass of my spacecraft.

In short, spacecraft don't re-enter the atmosphere at very shallow angles because of re-entry stress (although that is a concern,) they do it because it's simply the best they can do with the fuel they have onboard.

  • $\begingroup$ its not me its animations of the meteor that killed the dinosaurs $\endgroup$ Feb 9 '18 at 19:20
  • $\begingroup$ i wont forgive assumptions either btw :) $\endgroup$ Feb 9 '18 at 19:24
  • $\begingroup$ @user1886419 Joy of Tech $\endgroup$
    – Barmar
    Feb 9 '18 at 20:40

One possibility is to launch straight up. Read about sounding rockets and Blue Origin New Sheppard vehicle.

If you launch straight up and do not reach escape velocity, you will come almost straight down. Almost because the Earth rotates under you and when you leave the atmosphere you are no longer affected by that rotation. Then falling down you will reach velocity dependent on the maximal height achieved and may not burn if you did not fly too high.

  • 4
    $\begingroup$ Or you can launch from the pole. No velocity up there ;) $\endgroup$
    – Antzi
    Feb 9 '18 at 11:05
  • $\begingroup$ @Antzi You would be rotating though. Could you use that rotational energy in any way if you were performing a polar launch? $\endgroup$
    – asawyer
    Feb 9 '18 at 16:50
  • $\begingroup$ @asawyer one rotation a day is insignificant energy. $\endgroup$
    – Jeffrey
    Feb 9 '18 at 18:34
  • $\begingroup$ From what I can tell they don't ever leave earth. Is it possible though? $\endgroup$ Feb 9 '18 at 19:26
  • $\begingroup$ The rotational energy from the launch pad location is not set to zero by just leaving the atmosphere. The rotational energy is zero only when launching from the north or south pole. At other locations, some part of the propellant is needed to cancel the rotational energy and go up really straight up. $\endgroup$
    – Uwe
    Feb 9 '18 at 22:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.