Questions For Space Gamers

[high G]

Here is a question for the space gamers: You have a lander and you've
tasked it to set down on a small moon and take mining samples. This moon
has a proto-atmosphere. As the lander leaves orbit and approaches the
moon, you know the lander will fall towards the moon at a constant 1000 miles/hour
(the proto-atmosphere is enough for drag.) Luckily, you had a good year
so far and had the credits to buy a retro rocket which will decelerate
your lander at 20,000 miles/hour² (this is its acceleration-factor that
you purchased. It is constant, and the moon's gravitation acceleration
is included in this number. It's a good retro rocket.) So, seriously, at
what Altitude above the moon's surface should I fire the retro rocket so
it comes in for a smooth landing (that is, its velocity = 0 just as its
altitude = 0 at touchdown) ?

[Drejk]

25 miles

[Java Man]

Nuke the site from orbit, it's the only way to be sure.

[Tender Tendrils]

Can you spin the lander around so that firing the thruster accelerates you even faster?

[quibblemuch]

Java Man wrote:

Nuke the site from orbit, it's the only way to be sure.

Hold on a second, this installation has a substantial dollar value attached to it.

[MageHunter]

quibblemuch wrote:

Java Man wrote:

Nuke the site from orbit, it's the only way to be sure.

Hold on a second, this installation has a substantial dollar value attached to it.

I don't know... Multi-million dollar facility seems pretty cheap for something built on another planet.

[John Napier 698]

high G wrote:

Here is a question for the space gamers: You have a lander and you've

tasked it to set down on a small moon and take mining samples. This moon
has a proto-atmosphere. As the lander leaves orbit and approaches the
moon, you know the lander will fall towards the moon at a constant 1000 miles/hour
(the proto-atmosphere is enough for drag.) Luckily, you had a good year
so far and had the credits to buy a retro rocket which will decelerate
your lander at 20,000 miles/hour² (this is its acceleration-factor that
you purchased. It is constant, and the moon's gravitation acceleration
is included in this number. It's a good retro rocket.) So, seriously, at
what Altitude above the moon's surface should I fire the retro rocket so
it comes in for a smooth landing (that is, its velocity = 0 just as its
altitude = 0 at touchdown) ?

So, let's simplify this. The retro-rocket has an acceleration value of 20,000 miles / hour². this simplifies to 20,000(5,280)ft / 3600s² = 105,600,000 ft / 12,960,000s². Which equals 8.15 ft/s². So, working on the velocity of 1000 mph( or 14,667 ft/s ), a velocity of zero can be reached at t = 14667 / 8.15 = 180 seconds. This corresponds to the altitude of h = at² or 8.15( 180²) or 263,940 ft, or 50 miles.

[Java Man]

D = 1/2 at^2, so it should be 25 miles, not 50.

[The Sideromancer]

Who uses Imperial measurements in spacecraft? Haven't we learned from our mistakes?

[Drejk]

The Sideromancer wrote:

Who uses Imperial measurements in spacecraft? Haven't we learned from our mistakes?

If you use the formulas it doesn't really matter as long as you correctly process the formula and don't change the measurement in the mean time.

Though John needlessly changed the measurements used adding to the complexity of calculations instead of simplifying them though. It can be all calculated using miles per hour and representing time as a fraction of hour (1/20).

[Morkus from Orcus]

{attempts math, fails} Shazbot!

[John Napier 698]

Java Man wrote:

D = 1/2 at^2, so it should be 25 miles, not 50.

Correct. My mistake. Been a long time since my Physics course.

[Java Man]

John Napier 698 wrote:

Java Man wrote:

D = 1/2 at^2, so it should be 25 miles, not 50.

Correct. My mistake. Been a long time since my Physics course.

A long time for me too, but I've had a refresher tutoring my sprout (who's in high school now? I'm not old enough for that!)

[John Napier 698]

Single with no children. High School Physics was over 30 years ago. Maybe, I'll buy a used textbook and go through the exercises as a refresher.

[Java Man]

I've seen more folks learn/ refresh math and science for gaming purposes than nongamers would ever guess.

[high G]

NEW QUESTION: (need help)

Consider the vertical launch of a two stage vehicle (e.g. a rocket) with
a 1000kg payload and a first stage structural-plus-propellant mass of
100,000kg, β=0.04, I_sp=300s, and thrust of 1.5e6 N. The respective
second stage values are 5,000, 0.08, 350, and 180,000.

a. If gravity and drag are neglected, what is the burn out velocity? Is
this enough to escape the Earth?

b. Assume a uniform gravity field of 9.8 m/s^2 and no other external
forces. What is the burn out velocity? Is this enough to escape the Earth?

.

[Drejk]

Are you using this forum to make us help you escape the Earth?

...

Are you an alien stuck on Earth who is building your own rocket?

[high G]

*****

So, the next Mars Rover built at JPL lands on Mars this Thursday afternoon (18th).
It will hit the top of the Martian atmosphere going v=48,144mph and has a time to impact,
I mean time to landing, of 7 minutes. So, what kind of average acceleration will it under go?

[high G]

Drejk wrote:

Are you using this forum to make us help you escape the Earth?

...

Are you an alien stuck on Earth who is building your own rocket?

Yes to both. And "I'm still here" ( <- part one of the Beard Trilogy )

ok, so next question.

Do I subtract the 'circle from the ellipse' or the 'ellipse from the circle'?

And as an aside question, is the canonical variable name for the distance from the ellipse's center to Earth 'f' or 'c' ??

[AceofMoxen]

The Sideromancer wrote:

Who uses Imperial measurements in spacecraft? Haven't we learned from our mistakes?

$125 million dollars later...

quibblemuch wrote:

Java Man wrote:

Nuke the site from orbit, it's the only way to be sure.

Hold on a second, this installation has a substantial dollar value attached to it.

They can bill me.

[Horwierd Silmuc]

Nice try Jeff Bezos, nice try…