A de-orbit burn, similar to that presented in the previous math problem, has been performed. During this de-orbit burn a pre-calculated ∆V (delta V, change in velocity) of 290 ft/s (or 88.4 m/s) will be used to decrease the Shuttle’s altitude from 205 miles to 60 miles at perigee.The Shuttle’s Orbital Maneuvering System (OMS) engines provide a combined thrust of 12,000 force-pounds or 53,000 Newtons.The Shuttle weighs 2.50 x 105 lbs when fully loaded. The Shuttle has a mass of 1.13 x 105 kg when fully loaded.
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
Method 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.
To calculate the duration of the de-orbit burn, we can use either method 1 or method 2 provided in the question. Let's go through both methods using the given values:
Method 1:
We have the force (F) provided by the OMS engines, which is 12,000 force-pounds or 53,000 Newtons. The mass (M) of the shuttle when fully loaded is 1.13 x 10^5 kg. The change in velocity (∆V) is 88.4 m/s.
Using Newton's second law, F = ma, we can rearrange the equation to solve for acceleration (a):
a = F / M
Substituting the values, we get:
a = 53,000 N / 1.13 x 10^5 kg
Now, we can use the equation for average acceleration, a = ∆V / t, and rearrange it to solve for time (t):
t = ∆V / a
Substituting the values, we get:
t = 88.4 m/s / (53,000 N / 1.13 x 10^5 kg)
Calculating this, we find:
t ≈ 0.00185 seconds
Since we want the time in minutes and seconds, we need to convert this to minutes and seconds:
t_minutes = t / 60
t_seconds = t % 60
Converting, we get:
t_minutes ≈ 0.00185 seconds / 60 ≈ 0.000031 minutes
t_seconds ≈ 0.00185 seconds % 60 ≈ 0.111 seconds
Therefore, the de-orbit burn must last approximately 0.000031 minutes (or about 0.111 seconds).
Method 2:
In this method, we're given the weight (W) of the shuttle when fully loaded, which is 2.50 x 10^5 lbs. The change in velocity (∆V) is 290 ft/s, and the acceleration (a) is in units of g (32 ft/s^2).
Using the equation F = Wa, we can solve for acceleration (a):
a = F / W
Substituting the values, we get:
a = 12,000 force-pounds / 2.50 x 10^5 lbs
Now, we can use the equation for average acceleration, a = ∆V / t, and rearrange it to solve for time (t):
t = ∆V / a
Substituting the values, we get:
t = 290 ft/s / (12,000 force-pounds / 2.50 x 10^5 lbs)
Calculating this, we find:
t ≈ 0.00604 seconds
Again, we need to convert this to minutes and seconds:
t_minutes = t / 60
t_seconds = t % 60
Converting, we get:
t_minutes ≈ 0.00604 seconds / 60 ≈ 0.000101 minutes
t_seconds ≈ 0.00604 seconds % 60 ≈ 0.362 seconds
Therefore, the de-orbit burn must last approximately 0.000101 minutes (or about 0.362 seconds).
Please note that there may be slight variations in the final decimals due to rounding during calculations.