Question

What is the ratio of the gravitational pull of the sun on the moon to that of the earth on the moon? (Assume the distance of the moon from the sun can be approximated by the distance of the earth from the sun.) Use the data in Appendix F. Is it more accurate to say that the moon orbits the earth, or that the moon orbits the sun?

Short answer

The ratio is 2.253

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• Gravitational Pull of Sun
• Gravitational Pull of Moon

Significance of Newton’s Law of Gravitation

Determination of the ratio of the gravitational pull of sun on moon to that of earth on moon

The gravitational force of attraction given by the Newton’s law is given as,

$F=\frac{G{M}_1{M}_2}{{\left(d\right)}^2}$ …(i)

Here M₁ is the mass of the object (1), M₂ is the mass of the object (2), R₁is the Radius of object (1), R₂is the Radius of object (2)_, d is the distance between these two objects.

Let us consider the M₁ and M₂ as sun and moon respectively. The gravitational pull of sun on moon is given from equn (i) as,

$F_s=\frac{G{M}_1{M}_2}{{\left(d_s\right)}^2}$

Let us consider the M₁ and M₂ as earth and moon respectively. The gravitational pull of sun on moon is given from equn (i) as,

$F_{e-m}=\frac{G{M}_e{M}_m}{{\left(d_s\right)}^2}$

Here F_s-mis the sun-moon, F_e-mis the force of earth-moon, M_s is the mass of sun, M_m is the mass of the moon, M_e is the mass of the earth, d_s is the distance of sun to moon, d_e is distance of earth to moon.

The ratio is given as of the gravitational pull,

$\frac{F_{s-m}=\frac{G{M}_s{M}_m}{{\left(d_s\right)}^2}}{F_{e-m}=\frac{G{M}_e{M}_m}{{\left(d_s\right)}^2}}$

The common terms get cancelled each other and the above equation becomes,

$$\begin{gathered} \frac{F_{s-m}=\frac{M_s}{{\left(d_s\right)}^2}}{F_{e-m}=\frac{M_e}{{\left(d_s\right)}^2}} \\ \frac{F_{s-m}}{F_{e-m}}=\frac{M_s}{{\left(d_s\right)}^2}*\frac{{\left(d_e\right)}^2}{M_e} \end{gathered}$$

By substituting the values for M_s, d_e, d_s, M_e from the Appendix,

$$\begin{gathered} \frac{F_{s-m}}{F_{e-m}}=\frac{1.99*{10}^{30}}{{\left(1.5*{10}^8\right)}^2}*\frac{{\left(3.9*{10}^5\right)}^2}{5.97*{10}^{24}} \\ =\frac{30.2679*{10}^{40}}{13.4325*{10}^{40}} \\ \frac{F_{s-m}}{F_{e-m}}=2.253 \end{gathered}$$

The ratio of the gravitational pull of the sun on the moon to that of the earth on the moon is 2.253

The moon moves around the earth in 27 days, both the earth and moon rotates the sun.