Question

12. Taking the age of Earth to be about 4×10^9 years and assuming its orbital radius of 1.5 ×10^11 m has not changed and is circular, calculate the approximate total distance Earth has traveled since its birth (in a frame of reference stationary with respect to the Sun).

Answer 1

The approximate total distance Earth has traveled since its birth is 3.8×10²¹m.

The Earth moves in a circular orbit of radius r around the Sun and it takes 1 year to go once around the Sun. Thus in a year it travels a distance equal to 2πr, which is the circumference of the Earth's orbit.

Therefore in n years that have elapsed, it would travel a distance d given by,

$d=2\pi rn$

Substitute 3.14 for π, 1.5×10¹¹m for r and 4×10⁹ for n.

$d=2\pi rn\\= 2(3.14)(1.5*10^1^1m)(4*10^9)\\ =3.768*10^2^1m=3.8*10^2^1m$

Thus, the approximate total distance Earth has traveled since its birth is 3.8×10²¹m.

Answer 2

The approximate total distance Earth has traveled since its birth is about:

3.8 × 10²¹ m

$\texttt{ }$

Further explanation

Angular Speed can be formulated as follows:

$\large {\boxed {\omega = \frac{ v } { R } }$

ω = Angular Speed ( rad/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

$\texttt{ }$

Linear Speed can be formulated as follows:

$\large {\boxed {v = \frac{ 2 \pi R } { T } }$

T = Period of Circular Motion ( s )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

$\texttt{ }$

Given:

total time taken = t = 4 × 10⁹ years

period of Earth's revolution = T = 1 year

orbital radius = R = 1.5 × 10¹¹ m

Asked:

total distance = d = ?

Solution:

We will use following formula to solve this problem:

$\texttt{Total Distance = Number of Revolutions } \times \texttt{ Circumference of Earth's Orbit}$

$d = N \times ( 2 \pi R )$

$d = \frac{t}{T} \times ( 2 \pi R )$

$d = \frac{4 \times 10^9}{1} \times { 2 \pi \times 1.5 \times 10^{11} }$

$d = 4 \times 10^9 \times 2 \pi \times 1.5 \times 10^{11}$

$d = 1.2 \pi \times 10^{21} \texttt{ m}$

$d \approx 3.8 \times 10^{21} \texttt{ m}$

$\texttt{ }$

Learn more

• Impacts of Gravity : brainly.com/question/5330244
• Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
• The Acceleration Due To Gravity : brainly.com/question/4189441

$\texttt{ }$

Answer details

Grade: High School

Subject: Physics

Chapter: Circular Motion