Question

The distance of the earth from the sun is 93 000 000 miles. if there are 3.15 × 107 s in one year, find the speed of the earth in its orbit about the sun.

Answer 1

The angular velocity of the orbit about the sun is:

w = 1 rev / year = 1 rev / 3.15 × 10^7 s

 

Now in 1 rev there is 360° or 2π rad, therefore:

w = 2π rad / 3.15 × 10^7 s

 

To convert in linear velocity, multiply the rad /s by the radius:

v = (2π rad / 3.15 × 10^7 s) * 93,000,000 miles

v = 18.55 miles / s = 29.85 km / s

Answer 2

Answer:

The speed of the earth about the sun is 4730.15 m/s.

Explanation:

It is given that,

The distance of the earth from the sun is, $d=93000000\ miles=1.49\times 10^{11}\ m$

Time taken to move from the earth to the sun, $t=3.15\times 10^7\ s$

We need to find the speed of the earth in its orbit about the sun. Let it is given by v. The total distance covered divided by total time taken is called the speed of the earth. Mathematically, it is given by :

$v=\dfrac{d}{t}$

$v=\dfrac{1.49\times 10^{11}\ m}{3.15\times 10^7\ s}$

v = 4730.15 m/s

So, the speed of the earth in its orbit about the sun is 4730.15 m/s. Hence, this is the required solution.