Question

problem 8:
the distance from mercury to the sun is approximately ( r = 5.79\times10^{10}) m.
the mass of the sun is ( m_s = 2\times10^{30}) kg. find what one year in mercury is
in terms of earth days, ( t =?) make sure to solve the problem, starting from
newtons second law and the universal law of gravitation.

Explanation:

Step1: Equate gravitational force and centripetal force

According to Newton's second law \(F = ma\), for a planet in circular - orbit around the Sun, the gravitational force \(F_g=\frac{GM_sm}{R^{2}}\) (from the universal law of gravitation, where \(G = 6.67\times10^{-11}\space N\cdot m^{2}/kg^{2}\), \(M_s\) is the mass of the Sun, \(m\) is the mass of the planet, and \(R\) is the orbital radius) provides the centripetal force \(F_c = m\frac{v^{2}}{R}\). So, \(\frac{GM_sm}{R^{2}}=m\frac{v^{2}}{R}\). The mass of the planet \(m\) cancels out, and we get \(v=\sqrt{\frac{GM_s}{R}}\).

Step2: Express the orbital period formula

The orbital period \(T=\frac{2\pi R}{v}\). Substitute \(v=\sqrt{\frac{GM_s}{R}}\) into the formula for \(T\). Then \(T = 2\pi\sqrt{\frac{R^{3}}{GM_s}}\)

Step3: Substitute the given values

We know that \(R = 5.79\times 10^{10}\space m\), \(M_s=2\times 10^{30}\space kg\), and \(G = 6.67\times 10^{-11}\space N\cdot m^{2}/kg^{2}\)

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Step4: Convert seconds to Earth - days

Since there are \(86400\space s\) in a day (\(1\space day=24\times3600\space s\)), then \(T=\frac{2.39\times 10^{6}}{86400}\approx 88\space days\)

Answer:

One year on Mercury is approximately \(88\) Earth - days.