Question

The planets have angular momentum LA and LB about the star. The value of is : Consider two planets, A and B, of masses MA = 1.0 × 1024 kg and MB = 2.0 × 1024 kg. in circular orbits of radius RA = 1.0 × 1011 m and RB = 4 × 1011 m around a common star of mass MS = 24 × 1031 kg. The two planets are orbiting in the same direction. You may assume that the planets are point masses and have no rotation about their own individual axes. (Take : G = N-m2/kg2). Find La/Lb

Answer 1

1/4

Answer 2

The angular momentum of a point mass in a circular orbit about the center is given by $L = mvr$, where $m$ is the mass of the object, $v$ is its speed, and $r$ is the radius of the orbit.

For a planet in a circular orbit around a star of mass $M_S$, the gravitational force provides the centripetal force:

$\frac{G M_S m}{r^2} = \frac{mv^2}{r}$

Solving for the speed $v$:

$v^2 = \frac{G M_S}{r}$

$v = \sqrt{\frac{G M_S}{r}}$

Substitute this expression for $v$ into the angular momentum formula:

$L = m \left(\sqrt{\frac{G M_S}{r}}\right) r$

$L = m \sqrt{\frac{G M_S r^2}{r}}$

$L = m \sqrt{G M_S r}$

For planet A, the angular momentum $L_A$ is:

$L_A = M_A \sqrt{G M_S R_A}$

For planet B, the angular momentum $L_B$ is:

$L_B = M_B \sqrt{G M_S R_B}$

We need to find the ratio $\frac{L_A}{L_B}$:

$\frac{L_A}{L_B} = \frac{M_A \sqrt{G M_S R_A}}{M_B \sqrt{G M_S R_B}}$

$\frac{L_A}{L_B} = \frac{M_A}{M_B} \sqrt{\frac{G M_S R_A}{G M_S R_B}}$

Since $G$ and $M_S$ are constants for both planets, they cancel out in the ratio:

$\frac{L_A}{L_B} = \frac{M_A}{M_B} \sqrt{\frac{R_A}{R_B}}$

Now, substitute the given values:

$M_A = 1.0 \times 10^{24}$ kg $M_B = 2.0 \times 10^{24}$ kg $R_A = 1.0 \times 10^{11}$ m $R_B = 4.0 \times 10^{11}$ m

$\frac{L_A}{L_B} = \frac{1.0 \times 10^{24}}{2.0 \times 10^{24}} \sqrt{\frac{1.0 \times 10^{11}}{4.0 \times 10^{11}}}$

$\frac{L_A}{L_B} = \frac{1}{2} \sqrt{\frac{1}{4}}$

$\frac{L_A}{L_B} = \frac{1}{2} \times \frac{1}{\sqrt{4}}$

$\frac{L_A}{L_B} = \frac{1}{2} \times \frac{1}{2}$

$\frac{L_A}{L_B} = \frac{1}{4}$

Explanation:

The angular momentum of a planet in a circular orbit is $L = mvr$. The orbital speed $v$ is determined by the balance of gravitational and centripetal forces, $v = \sqrt{GM_S/r}$. Substituting $v$ into the expression for $L$ gives $L = m\sqrt{GM_S r}$. The ratio of angular momenta is then $\frac{L_A}{L_B} = \frac{M_A\sqrt{GM_S R_A}}{M_B\sqrt{GM_S R_B}} = \frac{M_A}{M_B}\sqrt{\frac{R_A}{R_B}}$. Substituting the given values for masses and radii yields the ratio $\frac{1}{2}\sqrt{\frac{1}{4}} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.