A de-orbit burn, similar (but not the same) 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 99.1 meters per second will be used to decrease the Shuttle’s altitude from 222.5 miles to 60 miles at perigee. The Shuttle’s Orbital Maneuvering System (OMS) engines provide a combined thrust force of 53,000 Newtons. The Shuttle had a mass of 113,000 kg when fully loaded.
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 222.5 miles to 60 miles at perigee with a ∆V of 99.1m/s. Use the equations and conversions provided below to find the required burn time.
Equation Method:
Newton’s Second Law: F = ma
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
Equation that defines average acceleration, the amount by which velocity will change in a given amount of time: a = ∆V/t
Rearranging the acceleration equation above to find the time required for a specific velocity change given a specific acceleration, where t = ∆V/a
∆V = change in velocity in meters per second (mps)
a = acceleration is in meters per second per second, m/s2
t = required time in seconds
My work:
f= 53,000 Newtons
m= 113000kg
a= 99.1 meters/second
Question is: How do I calculate it when the problem occurs that I have already solved Newton's second law, I am lost where to go after that.
That is not the correct value for a, you must divide f by m giving you .469m/s2
From there divide the original 99.1 m/s by .469m/s^2 to give you your time.
To calculate the required burn time for the de-orbit burn, you can use the equation that defines average acceleration, which is:
a = ∆V / t
Given that the ∆V (change in velocity) is 99.1 meters per second, you can substitute these values into the equation:
99.1 = a * t
Now, to find the time required for the specific velocity change, you need to rearrange the equation to solve for 't':
t = ∆V / a
Substituting the given values, you get:
t = 99.1 / a
To find 'a', you can use Newton's second law, which states:
F = ma
The force (F) in this case is the combined thrust force provided by the OMS engines, which is 53,000 Newtons. The mass (m) value is given as 113,000 kg. Substituting these values into the equation, you get:
53,000 = 113,000 * a
Solving for 'a':
a = 53,000 / 113,000
Now, substitute this value of 'a' back into the equation for 't':
t = 99.1 / (53,000 / 113,000)
Simplifying this equation gives you the required burn time in seconds. To convert it into minutes and seconds, divide the total time in seconds by 60 (since there are 60 seconds in a minute) to get the number of minutes. The remaining seconds will give you the final answer.
I hope this helps you with your calculations! Let me know if you have any further questions.
To calculate the duration of the de-orbit burn, you can use the equation that relates acceleration, change in velocity (∆V), and time (t):
a = ∆V / t
You already have the value for ∆V, which is 99.1 meters per second. Now, let's substitute the known values into the equation:
99.1 m/s = (53,000 N) / 113,000 kg × t
To find the value of time (t), we need to rearrange the equation. Divide both sides of the equation by 113,000 kg:
99.1 m/s = (53,000 N) / (113,000 kg × t)
To isolate t, divide both sides of the equation by (53,000 N):
(99.1 m/s) × (113,000 kg) = t
t ≈ 5.8062 seconds
So, the de-orbit burn should last approximately 5.8062 seconds to achieve the desired change in altitude. However, since the question asks for the duration in minutes and seconds, we need to convert this time to minutes. There are 60 seconds in a minute.
5.8062 seconds ≈ 0 minutes and 5.8062 seconds
Therefore, the de-orbit burn should last approximately 0 minutes and 5.8062 seconds.