# Effect of Centrifugal Force on Weight

Quote of the Day

If you are not annoying someone you are not doing anything new.

— Michael Stainer

## Introduction

Figure 1: North American X-15 Hypersonic
Rocket-Powered Aircraft.  (Source)

I love to read Quora, and I often see interesting factoids there that I inspire me to put pencil to paper and verify them. This week, I read a response to the question "What does the pilot of a supersonic fighter feel when flying at Mach 3 at 40,000 feet?" I found one of the answers particularly interesting because of how the respondent generalized the question to make it more interesting. I love when people take a basic question and turn it into a more interesting question.

The response that intrigued me began by pointing out that an airplane flying at 40,000 feet with a velocity of Mach 3 would probably be melting – the same answer given by other respondents. The author then changed the question by increasing the airplane's speed to Mach 8, its altitude to 70,000 feet, and its location to the equator. I quote part of his response here.

Actually, as the velocity increases, a pilot or passenger would feel slightly less weight, because the Earth's curvature becomes very important. This gets a little unusual because things start to matter that you would not expect are relevant. For example, if you are flying straight east over the equator at Mach 8, you would feel only 87% of your weight, but if you were flying west at the same point, you would feel 93% of your weight. The difference is due to the Earth's rotation.

Since only a few manned research aircraft have even come close to this speed (e.g. Figure 1), I do not expect to be able to perform an actual test. However, a little math will confirm his statement. Let's dig in …

## Background

### Definitions

Mach Number (symbol: M)
A dimensionless quantity representing a fluid's velocity relative to the speed of sound, i.e. $M = \frac{v_{Fluid}}{c_{Sound}}$, where vFluid in this case is the air speed of the aircraft, and cSound is the speed of sound at that altitude. (Source)
Hypersonic
In aerodynamics, a hypersonic speed is one that is highly supersonic. Since the 1970s, the term has generally been assumed to refer to speeds of Mach 5 and above. (Source)

### Approach

Here is my approach to duplicating the respondent's results:

• determine the velocity represented by Mach 8 at 70,000 feet.

Mach number varies with altitude, and we need to determine the velocity in m/s for an airplane at 70,0000 feet with a speed of Mach 8.

• compute the velocity of an object on the Earth's surface at the equator.

Even an object sitting stationary on the equator is actually being carried along by the rotation of the Earth. This movement causes the weight of an object to be measurably less at the equator than at the pole by the factor 288/289, which I will demonstrate in my analysis.

• compute the reduction in apparent gravity on an object due to centrifugal force on an airplane traveling easterly at Mach 8 and 70,000 feet.

When moving easterly, the airplane's velocity adds to the intrinsic velocity of an object being carried along with the Earth's rotation.

• compute the reduction in apparent gravity on an object due to centrifugal force on an airplane traveling westerly  at Mach 8 and 70,000 feet.

When moving westerly, the airplane's velocity subtracts from the intrinsic velocity of an object being carried along with the Earth's rotation.

• the weight reduction at the equator while flying at altitude is relative to the person's sea level weight at the pole.

At the poles, a person has no weight reduction due to centrifugal force.

## Analysis

Figure 2 shows my mathematical work that duplicates the results cited above (87% and 94%), with the final numbers marked with light-yellow highlights.

Figure 2: My Analysis of the Weight Reduction at the Equator.

## Conclusion

I was surprised that high-speed, high-altitude flight could experience enough centrifugal force to have  a measurable effect on the weight of a person. At first glance, I found the weight reductions stated to be almost unbelievable – however, a bit of physics shows that the numbers are correct.

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### 5 Responses to Effect of Centrifugal Force on Weight

1. Ronan Mandra says:

Hi, your weight reduction formula for easterly travel included a term of:
(V_Earth)^2/R_Earth. This is the Centrifugal acceleration of a point on the earth's surface which seems to have nothing to do with an aircraft flying at 8 Mach at 70,000 ft. What did I miss?

• mathscinotes says:

Hi Ronan,

That term should not be in the easterly travel equation. I have made the correction. Thanks for the help in getting things right!

mathscinotes

2. Ronan Mandra says:

HI, I sometime feel like that snarky kid that points out where the teacher has not crossed their letter T's or dotted their letter I's. I believe you updated your code but your Figure 2 has not been updated.

• mathscinotes says:

I think of you as a great help! The only folks that I struggle with are the flat earth people, who have descended on me in droves.

As far as Figure 2, you were correct. I am having an issue with WordPress. If I replace an image with an image of the same name but later date, WordPress is grabbing the first version. I am now going to give each image a rev number. Thank you for the help. It is greatly appreciated.

mathscinotes

3. Alastair McGowan says:

Great to see these calculations. I empathise with your attracting flat earth interest. My whole reason for coming here was the opposite - flying around the equator in a flat earth model would involve similar physics but laterally. I am keen to know how much a person would lean sideways (both East and Westerly) at 500mph (altitude not a factor here). For flat earth adherents to demonstrate using a pendulum and adjustments for track. Of course the resulting 'debate' could become immensely irritating, but for me as a psychologist also rather interesting.