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July 30, 2016
Posted by **Levi B.** on Friday, January 20, 2012 at 6:55pm.

A little background: What is the difference between the Shuttle’s mass and weight? An object’s mass does not change from place to place, but an object’s weight does change as it moves to a place with a different gravitational potential. For example, an object on the moon has the same mass it had while on the Earth but the object will weigh less on the moon due to the moon’s decreased gravitational potential.The shuttle always has the same mass but will weigh less while in orbit than it does while on Earth’s surface.

Calculate how long a de-orbit burn must last in minutes and seconds to achieve the Shuttle’s change in altitude from 205 miles to 60 miles at perigee.Use the equations and conversions provided below in method 1 or method 2 to find the required burn time.Please report your answer in minutes and seconds by converting your partial minutes into seconds.For example, if your answer is 92 seconds, convert to minutes by multiplying your answer by 1min/60 seconds to get 1.5333 minutes. Then take the 0.53333 minutes and convert into seconds by multiplying by 60 seconds per minute to get 32 seconds. Your reported quantity would be 1 minute and 32 seconds.

Method 1: F = ma Newton’s Second Law, where:

a, acceleration is in meters per second per second (m/s2)units

F, force is in Newtons (1 N = 1kg m/s2 )

M, mass is in kg units

a = ∆V/ t Equation that defines average acceleration, the amount by

which velocity will change in a given amount of time.

t = ∆V/a Rearranging the acceleration equation above to find the

time required for a specific velocity change given a

specific acceleration, where:

∆V is change in velocity in meters per second (mps)

a, acceleration is in meters per second per second, m/s2

t, required time in seconds

OR USE,

Method 2: F = Wa, where:

a, acceleration is in units of g’s (1g is equal to 32 feet

per second per second, ft/s2)

F, force is in force pound units

W, weight is in units of pounds

a = ∆V/ tEquation that defines average acceleration, the average

amount by which velocity will change in a given amount

of time.

t = ∆V/a Rearranging the acceleration equation above to find the

time required for a specific velocity change given a

specific acceleration, where:

∆V is change in velocity in feet per second (ft/s)

a, acceleration is in feet per second per second units

t, required time in seconds

- physics -
**Eric**, Tuesday, February 5, 2013 at 12:05amMethod 1

Step 1: Solve for acceleration

From F=ma, rearrange to a=F/m

a=F/m= (53000 N)/(1.13 x 〖10〗^5 kg)= .469026549 〖m/s〗^2

Step 2: Solve for time

t=∆v/a=(88.4 m/s)/(.469026549 〖m/s〗^2 )=188.4754716 seconds

Step 3: Convert seconds to minutes

(188.4754716 seconds)/60=3.141257859 minutes

Step 4: Convert minutes to seconds

.141257859 × 60=8.47547154 seconds

Final answer:

It would require 3 minutes and 8 seconds of burn time to change the Shuttle’s altitude from 205 miles to 60 miles at perigee.