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The mass of Earth is 5.972 x 1024 kg and its orbital radius is an average of 1.496 x 1011 m. Calculate its linear momentum, given the period of one rotation is 3.15 x 107 s

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nuhulawal20

Answer: its linear momentum is 1.78 × 10²⁹ kg.m/s

Explanation:

Given that;

mass of Earth m =  5.972 x 10²⁴ kg

radius r = 1.496 x 10¹¹ m

period t = 3.15 x 10⁷ s

now we know that Earth rotates in a circular path so the distance travelled per rotation is;

d = 2πr we substitute

d = 2π × 1.496 x 10¹¹ m

= 9.4 × 10¹¹ m

Now formula for speed v is;

v = d/t

we substitute

v = 9.4 × 10¹¹ m / 3.15 x 10⁷ s

v = 2.98 × 10⁴ m/s

now we determine the linear momentum p

linear momentum p = mv

we substitute

p = (5.972 x 10²⁴ kg) × (2.98 × 10⁴ m/s)

p = 1.78 × 10²⁹ kg.m/s

Therefore its linear momentum is 1.78 × 10²⁹ kg.m/s

raphealnwobi

The linear momentum of earth is 17.8 * 10²⁸ kgm/s

The orbital radius (r) of earth is 1.496 x 10¹¹ m. Hence the distance covered by one rotation is:

Distance = 2πr = 2π(1.496 x 10¹¹) m

The period of one rotation is 3.15 x 10⁷ s.

The velocity of earth (v) = distance/time = 2π(1.496 x 10¹¹) m/ 3.15 x 10⁷ s = 298840 m/s

Linear momentum = mass * velocity = 5.972 x 10²⁴ kg * 298840 m/s = 17.8 * 10²⁸ kgm/s

Therefore the linear momentum of earth is 17.8 * 10²⁸ kgm/s

Find out more at: https://brainly.com/question/12194595

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