Physics question I cannot figure out?

Question

the space shuttle performs a retro burn for 3.5min to go from an orbital velocity of 17500mph to a re-entry velocity of 5000mph.

1) how many miles does the maneuver require

2) how many G's do the astronauts experience.

I could figure this one

Answer 1

the first one I can help you answer only number two I am unsure of.

For number one all you need to use is the: speed = distance/time formuale

I don't have a calculator to hand but the calculation you would need to carry out would be:

17500=Distance/*whatever 3.5 mins is of an hour* (you will need a calculator to work that out!)

hope I helped!

xxxxx

Answer 2

17500 mph = 7823 m/s

5000 mph = 2235 m/s

3.5 min = 210 sec

Δv = at

5588 = a*210

a = 26.6 m/s²

a in g's = 26.6 / 9.8 = 2.7 Gs

for distance, I assume you want distance referenced to a spot on earth's surface.

d = ½at² + v₀t

d = -½(26.6)(210)² + 7823 *210

d = - 586530 + 1642830 m

d = 1056300 m or 656 miles

If you want distance referenced to the original orbit, than the answer is 586530 meters.

.

Answer 3

Supposed that the retro rocket exerts a constant retarding acceleration on the shuttle, the motion of the shuttle will be uniformly decelerated

a = change in v / time = (17500 -5000) /3.5 mph/minute

a = 12500 / 3.5*60 mile/minute/minute = 60 mile/minute/minute

(1)

(unit used: s in miles, t in minutes, and 'a' in mile/minute/minute)

s = (1/2) at^2 = 30 t^2 = 30 * (3.5)^2 = 735 miles

(2)

Converting retro acceleration 'a' to SI unit to compare with acceleration due to gravity G = 9.8 m/s/s

1 mile = 1,609 meters; 1 minute = 60 seconds

a = 30 miles/minute/minute = (30 * 1,609 / 60 /60) m/s/s = 13.4 m/s/s

Thus a/G = 13.4 / 9.8 = 1.4

or a = 1.4 G