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) ?

John Napier 698 |

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 includedin 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.

Drejk |

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).

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?

.