Question

The earth has a radius of 6380 km and turns around once on its axis in 24 h. (a) What is the radial acceleration of an object at the earth’s equator? Give your answer in m/s2 and as a fraction of g. (b) If arad at the equator is greater than g, objects will fly off the earth’s surface and into space. What would the period of the earth’s rotation have to be for this to occur?

Answer 1

Answer:

a) 0.03374 m/s²

b) 0.00343g

c) 5067 seconds or lower.

Explanation:

R = Radius of Earth = 6380000 m

T = Time period

$a_{rad}$ = Radial acceleration

From Kepler's third law

$a_{rad}=\frac{4\pi^2 R}{T^2}\\\Rightarrow a_{rad}=\frac{4\pi^2 6380000}{(24\times 60\times 60)^2}\\\Rightarrow a_{rad}=0.03374\ m/s^2$

g = Acceleration due to gravity = 9.81 m/s²

Radial acceleration is 0.03374 m/s²

1 g = 9.81 m/s²

$1\ m/s^2=\frac{1}{9.81}g$

$0.03374=\frac{0.03374}{9.81}g=0.00343g$

Radial acceleration is 0.00343g

b) Here, we have considered $a_{rad}=g$

$a_{rad}=\frac{4\pi^2 R}{T^2}\\\Rightarrow T=\sqrt{\frac{4\pi^2 R}{g}}\\\Rightarrow T=\sqrt{\frac{4\pi^2\times 6380000}{9.81}}\\\Rightarrow T=5067\ seconds$

Hence, the time period of the earth’s rotation would be 5067 seconds.

Increasing the radial acceleration would decrease the time period further.

Answer 2

Part A: The radial acceleration of the earth is 0.034 m/s2 or 0.0034 g.

Part B: The time period of the earth's rotation when the radial acceleration is greater than g is 1.4 hr.

How do you calculate the radial acceleration?

Given that the radius of the earth is 6380 km and the time interval in which the earth turns around once is 24 h.

Part A

The radial acceleration of the earth can be calculated by Kepler's third law.

$a = \dfrac {4\pi^2R}{t^2}$

Where a is the radial acceleration, R is the radius and t is the time interval.

$a = \dfrac { 4\times 3.14^2 \times 6380 \times 1000}{(24\times 60\times 60)^2}$

$a = 0.034 \;\rm m/s^2$

Acceleration due to gravity = 9.8 m/s² and radial acceleration is 0.034 m/s². This can be written as,

$1 g = 9.81\;\rm m/s^2$

$1m/s^2 = \dfrac {1}{9.8}g$

$0.034 \;\rm m/s^2 = \dfrac {1}{9.8}\times 0.034 g$

$0.034 \;\rm m/s^2 = 0.0034g$

Hence the radial acceleration of the earth is 0.034 m/s2 or 0.0034 g.

Part B

Given that the radial acceleration is greater than gravitational acceleration as the equator. Let us consider that,

a = g

Then the time period will be,

$T = \sqrt{\dfrac {4\pi^2R}{a}}$

$T = \sqrt{\dfrac {4\times 3.14^2\times 6380\times 1000}{9.8}}$

$T = 5067\;\rm s$

$T = 1.4 \;\rm hr$

Hence we can conclude that the time period of the earth's rotation when the radial acceleration is greater than g is 1.4 hr.

To know more about radial acceleration, follow the link given below.

https://brainly.com/question/7261283.