Newton's Law of GravitationGravitation

1301 Newton's Law of Gravitation

Definition

Every particle of matter in the universe attracks every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distant between them.

Fg=Gm1m2r2\begin{aligned} F_g = \frac{Gm_1m_2}{r^2} \end{aligned}

where GG is the Gravitational constant and rr is the distance between particles.

Exercises

1

(Quiz) Earth orbits sun at R=150R = 150 million km and travels 9393 million miles in 1 year(3.1×107s3.1\times 10^7s). Calculate the mass of the sun.

Solution

The attraction between Earth and Sun is the centripetal force that forces Earth to orbit.

Fg=FcGmemsR2=meve2r⇒ve=GmsR(1)\begin{aligned} F_g &= F_c\\ \frac{Gm_em_s}{R^2} &= \frac{m_ev_e^2}{r}\\ \To & v_e = \sqrt {\frac{Gm_s}{R}} &\text{(1)}\\ \end{aligned}

As Earth travels 9393 million miles in 1 year(3.1×107s3.1\times 10^7s)

ve=2πRt(2)\begin{aligned} v_e &= \frac{2\pi R}{t} &\text{(2)} \end{aligned}

Combine Eq.(1) and (2), we have

(R=1.5×1011mG=6.67×10−11m3/kg⋅s2t=3.1×107s)ms=4π2R3Gt2=4π2⋅(1.5×1011)36.67×10−11×(3.1×107)2=2.08×1030kg\begin{aligned} (R &= 1.5 \times 10^{11}m \quad G= 6.67 \times 10^{-11}m^3/kg\cdot s^2 \quad t=3.1\times 10^7s)\\ m_s &= \frac{4\pi^2 R^3}{Gt^2}\\ &=\frac{4\pi^2 \cdot (1.5\times 10^{11})^3}{6.67 \times 10^{-11} \times (3.1 \times 10^7)^2}\\ &= 2.08 \times 10^{30}kg \end{aligned}

(13.1) 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?

Solution

ratio = 2.18