# RPM required to simulate Earth's gravity (Centrifugal acceleration)

• jwlgrant
In summary, the problem is to determine how many RPMs a section of a spacecraft with a radius of 30 meters must rotate at to simulate one Earth gravity. One method used a formula of A = V^2/R and the other used ω = √(A/r) with RPM = 60ω/2∏ to solve for the answer. The correct answer is 5.5 RPM, not 14 RPM.
jwlgrant

## Homework Statement

Here's the problem: "On a journey to Mars, one design is to have a section of the spacecraft rotate to simulate gravity. If the radius of this section is 30 meters, how many RPMs must it rotate to simulate one Earth gravity (1 g = 9.8 meters/sec^2)?"

## Homework Equations

$A = \frac{V^2}{R}$

## The Attempt at a Solution

I completed this problem, getting an answer of 5.5 RPM. Apparently I was wrong, and the answer should have been 14 RPM. I asked my instructor about it, and he showed me the calculations used to get 14 RPM, but it still didn't quite make sense to me.

Here's the process my instructor used to solve the problem:
Circumference is C = 2∏(30) = 188 meters.

1 rpm is like rotating 1 full circular rotation every minute, so 188/60 = 3.1 meters/sec.

Then V = 3.1 meters/sec x RPM.

Then A = (3.1 RPM)^2/188 = 0.05 x rpm^2.

So: 9.8 = 0.05rpm^2

RPM = 14

--

Here's my process:
A = 9.8 (m)/(s^2)
R = 30 m
V = √(AR)
V = 17.15 m/s

1 revolution = circumference = 2∏30 m
[1 revolution/(2∏30 m)] * [60s/1 min] (1 rev is equal to 2π30m and sixty seconds is equal to one minute)
V = (17.15 m/s)/∏ RPM
V = 5.5 RPM

--

I realize, of course, that we both used different methods to get the answer. I thought I did it wrong, so I solved the problem again, using his method, to try to get the correct answer. He used 188m for the radius component when solving for A=V^2/R, but when I used 30m I got an answer of 5.5RPM. The radius, 30m, should be used in that equation, not 188m, right?

I would greatly appreciate any help in understanding this problem. I can't tell where/how I went wrong in solving it.

jwlgrant said:

## Homework Statement

Here's the problem: "On a journey to Mars, one design is to have a section of the spacecraft rotate to simulate gravity. If the radius of this section is 30 meters, how many RPMs must it rotate to simulate one Earth gravity (1 g = 9.8 meters/sec^2)?"

## Homework Equations

$A = \frac{V^2}{R}$

## The Attempt at a Solution

I completed this problem, getting an answer of 5.5 RPM. Apparently I was wrong, and the answer should have been 14 RPM. I asked my instructor about it, and he showed me the calculations used to get 14 RPM, but it still didn't quite make sense to me.

Here's the process my instructor used to solve the problem:
Circumference is C = 2∏(30) = 188 meters.

1 rpm is like rotating 1 full circular rotation every minute, so 188/60 = 3.1 meters/sec.

Then V = 3.1 meters/sec x RPM.

Then A = (3.1 RPM)^2/188 = 0.05 x rpm^2.
Why is he dividing by 188 (circumference)?? Should be 30.

So: 9.8 = 0.05rpm^2

RPM = 14

--

Here's my process:
A = 9.8 (m)/(s^2)
R = 30 m
V = √(AR)
V = 17.15 m/s

1 revolution = circumference = 2∏30 m
[1 revolution/(2∏30 m)] * [60s/1 min] (1 rev is equal to 2π30m and sixty seconds is equal to one minute)
V = (17.15 m/s)/∏ RPM
V = 5.5 RPM

--

I realize, of course, that we both used different methods to get the answer. I thought I did it wrong, so I solved the problem again, using his method, to try to get the correct answer. He used 188m for the radius component when solving for A=V^2/R, but when I used 30m I got an answer of 5.5RPM. The radius, 30m, should be used in that equation, not 188m, right?

I would greatly appreciate any help in understanding this problem. I can't tell where/how I went wrong in solving it.
You are right and the instructor is wrong.

It is easier to use Ac = ω2r; so $\omega = \sqrt{A/r}$ where ω is the angular speed in radians/sec.

Letting RPM = 60ω/2∏:

$RPM = \frac{60}{2\pi}*\sqrt{A/r} = \frac{60}{2\pi}\sqrt{9.8/30} = 5.5$

AM

Last edited:
Then A = (3.1 RPM)^2/188 = 0.05 x rpm^2.
I also agree, it must be divided by radius = 30 not circumference.

## 1. What is the formula for calculating the RPM required to simulate Earth's gravity?

The formula for calculating the RPM (revolutions per minute) required to simulate Earth's gravity is: RPM = √(g/0.00001118), where g is the desired acceleration in meters per second squared.

## 2. How does increasing the RPM affect the simulated gravity?

Increasing the RPM will increase the simulated gravity. This is because the centrifugal acceleration, which is caused by the spinning motion, increases with higher RPMs.

## 3. Is it possible to achieve exactly the same gravity as Earth's using RPM?

No, it is not possible to achieve exactly the same gravity as Earth's using RPM. This is because the centrifugal acceleration created by RPM is always slightly less than the gravitational acceleration on Earth.

## 4. What is the minimum RPM required to simulate Earth's gravity?

The minimum RPM required to simulate Earth's gravity depends on the radius of the rotating object. The smaller the radius, the higher the required RPM to achieve the same centrifugal acceleration.

## 5. Are there any real-world applications of simulating Earth's gravity using RPM?

Yes, there are real-world applications of simulating Earth's gravity using RPM. Some examples include centrifuges used in space to simulate gravity for experiments, and amusement park rides that use centrifugal force to create a sensation of weightlessness.

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