Question

Assuming that the earth momentum is at a distance 3.82*10^8m away from the centre of the earth, calculate it's linear speed and it's period of orbit motion round the earth.

Answer 1

Answer on Question #74585, Physics / Mechanics | Relativity

Assuming that the earth momentum is at a distance $3.82 \times 10^{8} \, \text{m}$ away from the centre of the earth, calculate its linear speed and its period of orbit motion round the earth.

Solution:

$$d = 3. 8 2 \times 1 0 ^ {8} m$$ $$M _ {\text {c e n t r e}} = M _ {\text {E r t h}} = 5. 9 8 \times 1 0 ^ {2 4} \mathrm {k g}$$ $$G = 6. 6 7 \times 1 0 ^ {- 1 1} N m ^ {2} / k g ^ {2}$$

We will begin by determining the orbital speed using the following equation:

$$v = \sqrt {\frac {G M _ {c e n}}{d}}$$ $$v = \sqrt {\frac {6 . 6 7 \times 1 0 ^ {- 1 1} \times 5 . 9 8 \times 1 0 ^ {2 4}}{3 . 8 2 \times 1 0 ^ {8}}} = 1. 0 2 \times 1 0 ^ {3} m / s$$

Finally, the period can be calculated using the following equation

$$\frac {T ^ {2}}{R ^ {3}} = \frac {4 \pi^ {2}}{G M _ {c e n}}$$

So,

$$T = \sqrt {\frac {4 \pi^ {2} \times (3 . 8 2 \times 1 0 ^ {8}) ^ {3}}{6 . 6 7 \times 1 0 ^ {- 1 1} \times 5 . 9 8 \times 1 0 ^ {2 4}}} = 2. 3 5 \times 1 0 ^ {6} s$$

Answer: $1.02 \times 10^{3} \frac{m}{s}$ and $2.35 \times 10^{6} s$

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