Question

There is a moon orbiting an Earth-like planet. The mass of the moon is 9.58 × 1022 kg, the center-to-center separation of the planet and the moon is 4.28 × 105 km, the orbital period of the moon is 28.9 days, and the radius of the moon is 1700 km. What is the angular momentum of the moon about the planet? Answer in units of kg m2 /s.

Answer 1

Answer:

= 4.38 × 10³⁴kgm²/s

Explanation:

Given that,

mass of moon m = 9.5 × 10²²kg

Orbital radius r = 4.28  × 10⁵km

Orbital period  T = 28.9days

T = 28.9  × 24 × 60 × 60

   = 2,496,960s

Angular momentum of the moon about the planet

L = mvr

L = mr²w

$L = mr^2\frac{2\pi }{T} \\\\L = \frac{9.5 \times 10^2^2 \times(4.28\times10^8)^2\times2\times3.14}{2496960} \\\\L = 4.389.5 \times 10^3^4kgm^2/s$

Answer 2

Answer:

2.4094 × $10^{14}$ kg $m^{2}$/sec

Explanation:

Mass of moon = 9.58 × $10^{22}$ kg

Radius of moon = 1700 km

Time period of moon = 28.9 days =2,496,960 sec

Angular momentum;

L=mvr   --- (1)

where v= velocity

Velocity of moon =2 π r / T

Put in above;

==> L = m (2 π r) r / T

==> L = 2 π m $r^{2}$ / T

Putting data values;

==> L = 2×3.14×9.58 × $10^{22}$ × 2890000000 ÷ 2,496,960

==> L = 2.4094×$10^{22}$×$10^{-8}$

==> L = 2.4094 × $10^{14}$ kg $m^{2}$/sec