Question

the ratio of Earth's orbital angular momentum to its mass is 4.4 into 10 power 15 m square second inverse the area enclosed by Earth's orbit is approximately

Answer 1

According to question,
$\bold{\frac{L_e}{M_e}=4.4 \times10^{15}}$
Here, Le is angular momentum and Me is the mass of earth.
we know,
Angular momentum = L = mωr²
here ω is angular velocity , r is the separation between observation point and axis of rotation { for earth , r = radius of it , when it rotates about axis passing through its diameter } and m is the mass of body

Now, ∵ ω = 2π/T , here T is time period
So, L = 2πmr/T

now, 2πMer²/MeT = 2πr²/T = 4.4 × 10¹⁵
Put T = 365 × 24 × 60 × 60 sec

⇒2πr²/365 × 24 × 60 × 60 = 4.4 × 10¹⁵
⇒2πr² = 4.4 × 10¹⁵ × 365 × 24 × 60 × 60
⇒πr² = 7 × 10²² m²

Hence, area enclosed by Earth's orbit is 7 × 10²² m²

Answer 2

Answer:

6.94×10^22 m^2

Explanation:

thanks