Calculate the centripetal acceleration of the Earth in its orbit around the Sun, and the net force exerted on the Earth. What exerts this force on the Earth? Assume that the Earth’s orbit is a circle of radius 1.50x10¹¹ m.

The relation between the force and acceleration of a rotating body is stated by Newton’s second law. The net force on a body will equal the mass of the body multiplied by the acceleration.

The radius of orbit is \(r = 1.50 \times {10^{11}}\;{\rm{m}}\).

The mass of Earth is \(m = 5.97 \times {10^{24}}\;{\rm{kg}}\).

The period of Earth in second is,

\(\begin{aligned}T &= 365\;{\rm{days}} \times \frac{{24\;{\rm{hrs}}}}{{1\;{\rm{days}}}} \times \frac{{3600\;{\rm{s}}}}{{1\;{\rm{hrs}}}}\\T &= 31536000\;{\rm{s}}\end{aligned}\)

The speed of Earth can be calculated as,

\(\begin{aligned}v &= \frac{{2\pi r}}{T}\\v &= \frac{{2\pi \left( {1.50 \times {{10}^{11}}\;{\rm{m}}} \right)}}{{31536000\;{\rm{s}}}}\\v &= 29.88 \times {10^3}\;{\rm{m/s}}\end{aligned}\)

The centripetal acceleration of Earth can be calculated as,

\(\begin{aligned}{a_c} &= \frac{{{v^2}}}{r}\\{a_c} &= \frac{{\left( {29.88 \times {{10}^3}\;{\rm{m/s}}} \right)}}{{1.50 \times {{10}^{11}}\;{\rm{m}}}}\\{a_c} &= 5.95 \times {10^{ - 3}}\;{\rm{m/}}{{\rm{s}}^2}\end{aligned}\)

Thus, the centripetal acceleration of Earth is \(5.95 \times {10^{ - 3}}\;{\rm{m/}}{{\rm{s}}^2}\).

The net force exerted on the Earth will be,

\(\begin{aligned}F &= m{a_c}\\F &= \left( {5.97 \times {{10}^{24}}\;{\rm{kg}}} \right)\left( {5.95 \times {{10}^{ - 3}}\;{\rm{m/}}{{\rm{s}}^2}} \right)\\F &= 3.55 \times {10^{22}}\;{\rm{N}}\end{aligned}\)

Thus, the net force on Earth is \(3.55 \times {10^{22}}\;{\rm{N}}\).