Question

Imagine that there is no force of gravity acting on you and you are at the equator at sea level (R = 6378km) . Using your answer to the previous question , calculate the centripetal force needed to keep you from flying off into space due to Earth's rotation if your mass is 50 kg .

Answer 1

The centripetal force required to keep a 50 kg person from flying off into space due to Earth's rotation at the equator is approximately 1.69 Newtons.

How to find centripetal force?

To calculate the centripetal force needed to keep a person from flying off into space due to Earth's rotation, we can use the formula for centripetal force:

`\[ F = m \cdot a_c \]`

where

F = centripetal force,

m = mass of the object (in this case, the person), and

`\( a_c \)` = centripetal acceleration.

Centripetal acceleration is given by:

`\[ a_c = (v^2)/(R) \]`

where

v = linear velocity at the equator due to Earth's rotation, and

R = radius of the Earth at the equator.

The linear velocity v can be calculated using the formula:

`\[ v = \omega \cdot R \]`

where

`\( \omega \)` = angular velocity of the Earth.

The Earth makes one complete rotation every 24 hours, so:

`\[ \omega = \frac{2\pi \text{ radians}}{24 \text{ hours}} \]`

Since

R = 6378 km, convert it to meters. Also, convert the time period to seconds to use in the formula.

R to meters: R = 6378 × 10³meters.

Time period to seconds: 24 hours = 24 × 60 × 60 seconds.

Calculate the centripetal force required:

Angular velocity:

`\[ \omega = (2\pi)/(24 * 60 * 60) \approx 7.2722 * 10^(-5) \, \text{radians per second} \]`

Linear velocity:

`\[ v = 7.2722 * 10^(-5) * 6378000 \approx 463.82 \, \text{m/s} \]`

Centripetal acceleration:

`\[ a_c = ((463.82)^2)/(6378000) \approx 0.03373 \, \text{m/s}^2 \]`

Centripetal force:

`\[ F = 50 * 0.03373 \approx 1.69 \, \text{Newtons} \]`

Therefore, the centripetal force required to keep a 50 kg person from flying off into space due to Earth's rotation at the equator is approximately 1.69 Newtons.