Hello and welcome to a very mathematical video! Let’s see if you can survive till the end. So,
here’s some backstory. You see, many months ago, I thought, a rocket launches itself using fuel.
Intuitively, if more fuel is added to the rocket, the higher it will go. However, one thing that many
people miss out on is the fact that, as a rocket’s fuel mass increases, so does its overall mass. As a
result, it will experience a stronger pull from the force of gravity, making its maximum height
achieved, lower.
So, now I come to the point of the video, I thought, is there a mathematical way to figure out the
point at which if you add more mass to a rocket, it’s maximum height will decrease. In another
words, is there a mathematical expression for “the optimal initial mass of a, very ideal rocket, for
it to reach the maximum height before its fuel runs out"? Well, let’s have a look!
Now, for the sake of this video, we will be adding the following assumptions, limitations and
simplifications. These of which are,
-
Linear mass loss, not realistic at all
-
No air, not realistic at all
-
Treating the rocket as a point, not realistic at all
-
Assuming 𝑔 is constant, not realistic at all
-
The rocket leaves perpendicular to the ground, not realistic at all
-
And you get,
Basically, it’s a very ideal rocket in which no physical phenomena such as temperature loss or
gain in the body which may somehow affect the performance of the boosters, even if that’s a
thing. An initial state which I’m going to assign to the rocket is that it’s initial velocity and
vertical displacement is 0 when the time is 0.
So, let me just explain the situation more visually…
Newton’s 2nd Law:
Σ𝐹 =
𝑑𝑝
𝑑𝑡
𝑝 = 𝑚𝑣:
Σ𝐹 =
𝑑(𝑚𝑣) 𝑑𝑚
𝑑𝑣
=
𝑣+𝑚
𝑑𝑡
𝑑𝑡
𝑑𝑡
Σ𝐹 = 𝑇 − 𝑚𝑔, and linear mass loss:
𝑇 − 𝑚𝑔 =
𝑑𝑚
𝑑𝑣
𝑣+𝑚
𝑑𝑡
𝑑𝑡
𝑚 = 𝑚0 − 𝜆𝑡
𝑑𝑣
𝑑(𝑚0 + 𝜆𝑡)
(𝑚0 + 𝜆𝑡) +
𝑣 = 𝑇 − (𝑚0 + 𝜆𝑡)𝑔
𝑑𝑡
𝑑𝑡
𝑑𝑣
(𝑚0 + 𝜆𝑡) + 𝜆𝑣 = 𝑇 − (𝑚0 + 𝜆𝑡)𝑔
𝑑𝑡
𝑑𝑣
𝜆
𝑇 − (𝑚0 + 𝜆𝑡)𝑔
+
𝑣=
𝑑𝑡 𝑚0 + 𝜆𝑡
𝑚0 + 𝜆𝑡
Integrating Factor:
𝑒
𝜆
∫𝑚 +𝜆𝑡𝑑𝑡
0
= 𝑒 ln(𝑚0+𝜆𝑡) = 𝑚0 + 𝜆𝑡
Multiply by the integrating Factor:
𝑑𝑣
(𝑚0 + 𝜆𝑡) + 𝜆𝑣 = 𝑇 − (𝑚0 + 𝜆𝑡)𝑔
𝑑𝑡
Reverse product rule on LHS:
𝑑
((𝑚0 + 𝜆𝑡)𝑣) = 𝑇 − (𝑚0 + 𝜆𝑡)𝑔
𝑑𝑡
Integrate both sides with respect to 𝑡:
𝑣(𝑚0 + 𝜆𝑡) = ∫(𝑇 − (𝑚0 + 𝜆𝑡)𝑔) 𝑑𝑡
𝑣(𝑚0 + 𝜆𝑡) = 𝑇𝑡 − 𝑔 ∫(𝑚0 + 𝜆𝑡) 𝑑𝑡
1
𝑣(𝑚0 + 𝜆𝑡) = 𝑇𝑡 − 𝑔𝑚0 𝑡 + 𝑔𝜆𝑡 2 + 𝑐
2
At 𝑡 = 0, 𝑣 = 0, so, 𝑐 = 0.
𝑣=
1
𝑇𝑡 − 𝑔𝑚0 𝑡 + 2 𝑔𝜆𝑡 2
𝑚0 + 𝜆𝑡
𝑑
Now, 𝑣 = 𝑑𝑡 𝑠
1
𝑇𝑡 − 𝑔𝑚0 𝑡 + 2 𝑔𝜆𝑡 2
𝑑
𝑠=
𝑑𝑡
𝑚0 + 𝜆𝑡
𝑠=∫
𝑠=
1
𝑇𝑡 − 𝑔𝑚0 𝑡 + 2 𝑔𝜆𝑡 2
𝑚0 + 𝜆𝑡
𝑑𝑡
(3𝑔𝑚02 − 2𝑇𝑚0 ) ln(𝜆𝑡 + 𝑚0 ) 𝑔𝜆𝑡 2 + (4𝑇 − 6𝑔𝑚0 )𝑡
+
+𝑐
2𝜆2
4𝜆
At 𝑡 = 0, 𝑠 = 0, so,
0=
(3𝑔𝑚02 − 2𝑇𝑚0 ) ln(𝑚0 )
+𝑐
2𝜆2
−(3𝑔𝑚02 − 2𝑇𝑚0 ) ln(𝑚0 )
𝑐=
2𝜆2
So,
𝑠=
(3𝑔𝑚02 − 2𝑇𝑚0 ) ln(𝜆𝑡 + 𝑚0 ) 𝑔𝜆𝑡 2 + (4𝑇 − 6𝑔𝑚0 )𝑡 (3𝑔𝑚02 − 2𝑇𝑚0 ) ln(𝑚0 )
+
−
2𝜆2
4𝜆
2𝜆2
𝝀𝒕 + 𝒎𝟎
(𝟑𝒈𝒎𝟐𝟎 − 𝟐𝑻𝒎𝟎 ) 𝐥𝐧 (
𝒎𝟎 ) 𝒈𝝀𝒕𝟐 + (𝟒𝑻 − 𝟔𝒈𝒎𝟎 )𝒕
𝒔=
+
𝟐𝝀𝟐
𝟒𝝀
Let 𝑠𝑚𝑎𝑥 be the maximum height attained by a rocket of initial mass, 𝑚0 . And, let 𝑡𝑒𝑥 be the time
when the rocket attains the maximum height, 𝑠𝑚𝑎𝑥 because of complete fuel exhaustion. Then,
(3𝑔𝑚02 − 2𝑇𝑚0 ) ln (
𝑠𝑚𝑎𝑥 =
𝜆𝑡𝑒𝑥 + 𝑚0
)
𝑚0
2𝜆2
Now, 𝑚 = 𝑚0 + 𝜆𝑡. So, 𝑡 =
𝑚−𝑚0
𝜆
. Hence, 𝑡𝑒𝑥 =
+
𝑚0 𝛼−𝑚0
𝜆
2
𝑔𝜆𝑡𝑒𝑥
+ (4𝑇 − 6𝑔𝑚0 )𝑡𝑒𝑥
4𝜆
where 𝑚0 𝛼 is the mass of the rocket
body without fuel and 0 < 𝛼 < 1.
(3𝑔𝑚02
𝑚 𝛼 − 𝑚0
𝜆( 0
) + 𝑚0
𝜆
− 2𝑇𝑚0 ) ln (
)
𝑚
0
𝑠𝑚𝑎𝑥 =
2𝜆2
𝑠𝑚𝑎𝑥 =
𝒔𝒎𝒂𝒙 =
(3𝑔𝑚02
𝑚 𝛼 − 𝑚0 2
𝑚 𝛼 − 𝑚0
𝑔𝜆 ( 0
) + (4𝑇 − 6𝑔𝑚0 ) ( 0 𝜆
)
𝜆
+
4𝜆
2
𝛼−1
2 𝑎−1
𝑔𝜆𝑚
+ (4𝑇𝑚0 − 6𝑔𝑚02 ) (
(
)
)
− 2𝑇𝑚0 ) ln(𝛼)
0
𝜆
𝜆
+
2𝜆2
4𝜆
(𝟑𝒈𝒎𝟐𝟎 − 𝟐𝑻𝒎𝟎 ) 𝐥𝐧(𝜶) 𝒈𝒎𝟐𝟎 (𝒂 − 𝟏)𝟐 + (𝟒𝑻𝒎𝟎 − 𝟔𝒈𝒎𝟐𝟎 )(𝜶 − 𝟏)
+
𝟐𝝀𝟐
𝟒𝝀𝟐
Now, we differentiate this equation with respect to 𝑚0 to find the vertex.
𝑑𝑠𝑚𝑎𝑥 (6𝑔𝑚0 − 2𝑇) ln(𝛼) 2𝑔𝑚0 (𝑎 − 1)2 + (4𝑇 − 12𝑔𝑚0 )(𝛼 − 1)
=
+
𝑑𝑚0
2𝜆2
4𝜆2
𝑑𝑠𝑚𝑎𝑥 (6𝑔𝑚0 − 2𝑇) ln(𝛼) + 𝑔𝑚0 (𝑎 − 1)2 + (2𝑇 − 6𝑔𝑚0 )(𝛼 − 1)
=
𝑑𝑚0
2𝜆2
𝑑𝑠𝑚𝑎𝑥 𝑚0 (6𝑔𝑙𝑛(𝛼) + 𝑔(𝛼 − 1)2 − 6𝑔(𝛼 − 1)) − 2𝑇𝑙𝑛(𝛼) + 2𝑇(𝛼 − 1)
=
𝑑𝑚0
2𝜆2
𝑑𝑠𝑚𝑎𝑥 𝑚0 (6𝑔𝑙𝑛(𝛼) + 𝑔(𝛼 − 1)2 − 6𝑔(𝛼 − 1)) − 2𝑇𝑙𝑛(𝛼) + 2𝑇(𝛼 − 1)
=
𝑑𝑚0
2𝜆2
Vertex occurs when
0=
𝑑𝑠𝑚𝑎𝑥
𝑑𝑚0
= 0,
𝑚0 (6𝑔𝑙𝑛(𝛼) + 𝑔(𝛼 − 1)2 − 6𝑔(𝛼 − 1)) − 2𝑇𝑙𝑛(𝛼) + 2𝑇(𝛼 − 1)
2𝜆2
2𝑇𝑙𝑛(𝛼) − 2𝑇(𝛼 − 1) = 𝑚0 (6𝑔𝑙𝑛(𝛼) + 𝑔(𝛼 − 1)2 − 6𝑔(𝛼 − 1))
𝑚0 =
2𝑇𝑙𝑛(𝛼) − 2𝑇(𝛼 − 1)
𝑇
2(𝑙𝑛(𝛼) − 𝛼 + 1)
= ×
2
6𝑔𝑙𝑛(𝛼) + 𝑔(𝛼 − 1) − 6𝑔(𝛼 − 1) 𝑔 6 ln(𝛼) + (𝛼 − 1)2 − 6(𝛼 − 1)
𝑇
𝑔
Let 𝑚𝑜𝑝𝑡 be the optimal initial mass. Let the constant after be 𝐶.
𝑪=
𝟐(𝒍𝒏(𝜶) − 𝜶 + 𝟏)
𝟔 𝐥𝐧(𝜶) + (𝜶 − 𝟏)𝟐 − 𝟔(𝜶 − 𝟏)
𝒎𝒐𝒑𝒕 =
𝑻
𝑪
𝒈
This is the optimal initial mass for the farthest height attained by a rocket. Note the rational
relationship here,
𝑚𝑜𝑝𝑡 𝑇
=
𝐶
𝑔