If the rotation of a planet of radius 3.41 × 106 m and free-fall acceleration 8.71 m/s2 increased to the point that the cen- tripetal acceleration was equal to the gravita- tional acceleration at the equator, what would be the tangential speed of a person standing at the equator?
How long would a day be in this case?
To find the tangential speed of a person standing at the equator, we need to calculate the centripetal acceleration first.
The centripetal acceleration is given by the formula:
ac = (v^2) / r
Where ac is the centripetal acceleration, v is the tangential speed, and r is the radius of the planet.
In this case, we have the centripetal acceleration equal to the gravitational acceleration at the equator, which is 8.71 m/s^2.
Now let's calculate the tangential speed:
8.71 = (v^2) / (3.41 × 10^6)
To find the tangential speed (v), we can rearrange the equation:
v^2 = 8.71 * (3.41 × 10^6)
v^2 = 29.7711 × 10^6
v = sqrt(29.7711 × 10^6)
v ≈ 5461.75 m/s
So, the tangential speed of a person standing at the equator would be approximately 5461.75 m/s.
To find the length of a day in this case, we need to consider that the planet rotates once in a day. The time it takes for a full rotation is equal to the circumference of the planet divided by the tangential speed.
The circumference of a circle is given by the formula:
C = 2πr
Where C is the circumference and r is the radius of the planet.
The length of a day is:
Day = (2πr) / v
Plugging in the given values:
Day = (2π * 3.41 × 10^6) / 5461.75
Day ≈ 3.1397 × 10^3 seconds
So, the length of a day in this case would be approximately 3.1397 × 10^3 seconds.