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
[tex]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[/tex]
Answer:
2.4094 × [tex]10^{14}[/tex] kg [tex]m^{2}[/tex]/sec
Explanation:
Mass of moon = 9.58 × [tex]10^{22}[/tex] 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 [tex]r^{2}[/tex] / T
Putting data values;
==> L = 2×3.14×9.58 × [tex]10^{22}[/tex] × 2890000000 ÷ 2,496,960
==> L = 2.4094×[tex]10^{22}[/tex]×[tex]10^{-8}[/tex]
==> L = 2.4094 × [tex]10^{14}[/tex] kg [tex]m^{2}[/tex]/sec
