# Force of a Superhero: Superman Returns

## Abstract

The purpose of this analysis is to determine the force Superman exerts to stop a plane from crashing into the ground.

## Introduction

In *Superman Returns*, there is a scene where Lois Lane is covering a story on a shuttle launch. She
is abroad a Boeing 777 plane when suddenly a malfunction occurs. Strapped to
the top of the plane is a shuttle orbiter that fails to disengage. The shuttle
orbiter launches but pulls the plane along with it. Coming to the rescue, Superman
projects his heat vision at the coupling system to release the shuttle orbiter.
Afterwards, Superman flies the shuttle orbiter away to safety. However, the
shuttle orbiter has burned off the tail of the plane, which sends the plane
into a downward spiral. Superman tries to grab one of the wings of the plane,
but it tears off. The second wing of the plane also tears off because of air
resistance. Finally, Superman flies in front of the plane and pushes up against
the nose of the plane to try to stop the plane from crashing into the ground.

To determine the force Superman exerts when he pushes up against the nose of the plane, I need to use the formulas:

a = (v–v_{0})/t

Where

a is the acceleration of the plane,

v is the final speed of the plane,

v_{0} is the initial speed of the plane, and

t is the time over which the object is accelerated

and …

ΣF = ma

Where

m is the mass of the plane and its passengers,

a is the acceleration of the plane, and

Σ F is the difference between the downward weight of the plane and the upward force of Superman

Thus …

ΣF = F_{superman}–W = ma

## Procedure

- View the scene here on YouTube. Locate the part where the plane's two wings have already been torn off.
- Using a stopwatch, measure the time from the moment Superman is pushing up against the nose of the plane until the moment he stops the plane. Do ten trials, and find the average time of these trials.

## Analysis

Trial # | Time (s) |
---|---|

1 | 29.74 |

2 | 29.38 |

3 | 29.16 |

4 | 29.98 |

5 | 30.06 |

6 | 29.79 |

7 | 29.47 |

8 | 29.65 |

9 | 29.88 |

10 | 30.10 |

In the movie, it was stated that Lois was traveling abroad a Boeing 777. According to the Aerospaceweb, the Boeing 777 can travel at 575 mph or 257 m/s. However, in the movie, since the plane loses both of its wings and is basically falling, I can assume that the actual speed of the plane is about 600 mph or 268 m/s. This value, 268 m/s, is the initial speed of the plane. The final speed of the plane is 0 m/s because this is when the plane comes to a stop. In addition, I found the time it took for Superman to stop the plane from crashing. The results are shown in the table on the right. The average time of these ten trials was 29.721 s.

With these values, I can first determine the acceleration of the plane over the time that it was stopped by using the equation

a = (v–v_{0})/t

Substituting the known values into the equation, I get:

a = (0 m/s–268 m/s)/29.721 s = –9.02 m/s^{2}

The next step is to figure out the mass of the plane and its passengers. According to the Aerospaceweb, when a Boeing 777-200 is empty, it has a mass of 135,580 kg. At maximum takeoff speed, a Boeing 777-200 has a mass of 247,210 kg. I will assume that the total mass of the plane and its passengers is 200,000 kg, a value between those two numbers.

Now I can finally calculate the force of Superman. As previously stated

ΣF = F_{superman}–W = ma

To find the force of Superman, I can rearrange the equation to get:

F_{superman} = W_{} + ma = m (g + a)

Substituting the known values into the equation, I get:

F_{superman} = (200,000 kg) (9.81 m/s^{2} + 9.02 m/s^{2}) = 3,766,000 N

## Conclusion

Therefore, the force Superman exerts to stop a plane from crashing into the ground is 3,766,000 N, which is an incredibly large amount of force.

## Sources of Error

- One source of error was that I estimated the values for the initial speed of the plane and the mass of the plane and its passengers.
- A second source of error was that my reaction time varied when I was using the stopwatch.

# Bonus Analysis 1: Torque Applied to the Plane

Another neat thing to analyze is the torque Superman applies to the plane after gently lowering it to the ground.

Assume the plane has its center of mass at its geometric center. Use the equations …

F = mg | and | r = length/2 |

where m is the mass of the plane and length is the length of the plane. As previously stated, the total mass of the plane and its passengers is 200,000 kg. According to the Aerospaceweb, the Boeing 777-200 also has a length of 63.73 m. Substituting the known values into the equation, I get:

F = mg = (200,000 kg) (9.81 m/s^{2}) = 1,962,000 N

r = length/2 = (63.73 m)/2 = 32 m

τ = Fr

τ = (1,962,000 N) (32 m) = 62,784,000 N/m

Therefore, the torque Superman applies to the plane after gently lowering it to the ground is 62,784,000 N/m, which is also an incredibly large amount of torque.

# Bonus Analysis 2: Superman's Free Fall

In *Superman Returns*, there is a scene where Lex Luthor stabs Superman
with a piece of Kryptonite. Superman gets up, takes a few stops back and falls down a cliff. The purpose
of this brief analysis is to determine the speed of Superman after he falls down a cliff.

The first method to determine the speed of Superman after he falls down the cliff is to use the kinematic equation

v = v_{0} + at

Where

v_{0} is the initial velocity of Superman,

a is the acceleration due to gravity 9.81 m/s^{2}, and

t is the time it takes for Superman to fall down the cliff

Trial # | Time (s) |
---|---|

1 | 6.52 |

2 | 6.43 |

3 | 6.07 |

4 | 6.39 |

5 | 6.52 |

6 | 6.12 |

7 | 6.84 |

8 | 6.30 |

9 | 6.52 |

10 | 6.16 |

Using a stopwatch, I measured the time from the moment Superman falls off the cliff until the moment he hits the water. I did ten trials and found the average time of these trials. The results are summarized in the table to the right. The average time of these ten trials was 6.387 s.

Now substituting the known values into the equation, I get:

v = 0 m/s + (9.81 m/s^{2}) (6.387 s) = 63 m/s

The second method to determine the speed of Superman after he falls down the cliff is to first find the height of the cliff. I can use the kinematic equation

s = v_{0}t + ½at^{2}

Where

s is the height of the cliff,

v_{0} is 0 m/s because Superman is not initially moving,

t is the time it takes for Superman to fall down the cliff, and

a is 9.81 m/s^{2}.

Substituting the known values into the equation, I get:

s = (0 m/s) (6.387 s) + ½ (9.81 m/s^{2}) (6.387 s)^{2} = 200 m

After calculating the height of the cliff, I can finally determine Superman's speed after he falls down the cliff using the equation

v^{2} = v_{0}^{2} + 2as

Where

v is the speed of Superman after he falls down the cliff,

v_{0} is 0 m/s because Superman is not initially moving,

a is 9.81 m/s^{2} and s is the height of the cliff.

Substituting the known values into the equation, I get:

v^{2} = (0 m/s)^{2} + 2 (9.81 m/s^{2}) (200 m) = 63 m/s

Therefore, using both methods, Superman's speed is 63 m/s after he falls down the cliff. However, there was one source of error during my analysis. I had neglected air resistance, which would make Superman's true speed somewhat less than 63 m/s.

Karen Fan -- 2005

Physics on Film- Feature Films
- Video Clips
- Video Games