Question

Light takes 8 minutes to reach the Earth, and the speed of light is 3.0×10^8 m/s. a) What is the orbital speed of the Earth around the Sun? b) What is its kinetic energy?

Answer 1

Answer:

(a) 28690 m/s (b) $2.46x10^{33}J$

Explanation:

The orbital speed is define as:

$v = \frac{2 \pi r}{T}$   (1)

Where r is the radius of the trajectory and T is the orbital period.

To determine the orbital speed of the Earth it is necessary to know the orbital period and the radius of the trajectory. That can be done by means of the Kepler's third law and average velocity equation.

The average velocity in a Uniform Rectilinear Motion is defined as:

$v = \frac{d}{t}$   (2)

Where v is the velocity, d is the covered distance and t is the time.

Equation 2 can be rewritten for d to get:

$d = vt$   (3)

In this case, v will be the speed of light and t, the 8 minutes that takes to reach the Earth.

The time will be converted to seconds so the units in equation 3 can match:

$8min . \frac{60s}{1min}$ ⇒  $480s$

$t = 480s$

Replacing all those values in equation 3 it is gotten:

$d = (3.0x10^{8}m/s)(480s)$

$d = 1.44x10^{11}m$

Kepler’s third law is defined as:

$T^{2} = r^{3}$

Where T is orbital period and r is the radius of the trajectory.

$T = \sqrt{r^{3}}$

$T = \sqrt{(1.44x10^{11}m)^{3}}$

It is necessary to pass from meters to astronomical unit (AU), 1 AU is defined as the distance between the Earth and the Sun.

$T = \sqrt{1AU}$

$T = 1AU$

That can be expressed in units of years.

$T = 1AU . \frac{1year}{1AU}$

$T = 1year$

But there are 31536000 seconds in one year:

$T = 1year . \frac{31536000s}{1year}$

$T = 31536000s$

Finally, equation  1 can be used:

$v = \frac{2 \pi (1.44x10^{11}m)}{(31536000s)}$

$v = 28690 m/s$

So Earth orbital speed around the Sun is 28690 m/s.

b) What is its kinetic energy?

The kinetic energy is defined as:

$E = \frac{1}{2}mv^{2}$  (4)

Notice that it is necessary to found the mass of the Earth, that can be done combining the Universal law of gravity and Newton's second law:

$F = \frac{GMm}{r^{2}}$

$ma = \frac{GMm}{r^{2}}$  (5)

M will be isolated in equation 5:

$M = \frac{r^{2}a}{G}$

Where r is the radius of the Earth ($6.38x10^{6}m$)

$M = \frac{(6.38x10^{6}m)^{2}(9.8m/s^{2})}{6.67x10^{-11}kg.m/s^{2}.m^{2}/Kg^{2}})$

$M = 5.98x10^{24} Kg$

$E = \frac{1}{2}( 5.98x10^{24} Kg))(28.690m/s)^{2}$

$E = 2.46x10^{33}Kg.m^{2}/s^{2}$

$E = 2.46x10^{33}J$

Hence, the kinetic energy of Earth is $2.46x10^{33}J$.