Question

1. a satellite is in orbit around the earth $1.44\times10^{8}$ m above the surface. the radius of the earth is $6.38\times10^{6}$ m and the mass of earth is $5.97\times10^{24}$ kg

a. what is the orbital speed of the satellite?
b. what is the orbital period of the satellite?

1. the distance from the center of the earth to the center of the moon is $3.84\times10^{8}$ m. if the orbital period is 27.3 days which is 2,358,720 seconds, determine:

a. the orbital speed of the moon.
b. the centripetal acceleration of the moon.

Explanation:

Problem 8

Step1: Calculate orbital radius

Orbital radius \( r = \text{Earth radius} + \text{height above surface} \)
\( r = 6.38 \times 10^6\ \text{m} + 1.44 \times 10^6\ \text{m} = 7.82 \times 10^6\ \text{m} \)

Step2: Orbital speed formula setup

Use gravitational centripetal force equilibrium: \( v = \sqrt{\frac{GM}{r}} \), where \( G=6.67 \times 10^{-11}\ \text{N·m}^2/\text{kg}^2 \), \( M=5.97 \times 10^{24}\ \text{kg} \)
\( v = \sqrt{\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{7.82 \times 10^6}} \)

Step3: Compute orbital speed

First calculate numerator: \( 6.67 \times 10^{-11} \times 5.97 \times 10^{24} \approx 3.98 \times 10^{14} \)
Then divide by \( r \): \( \frac{3.98 \times 10^{14}}{7.82 \times 10^6} \approx 5.09 \times 10^7 \)
Take square root: \( v = \sqrt{5.09 \times 10^7} \approx 7.13 \times 10^3\ \text{m/s} \)

Step4: Orbital period formula setup

Use \( T = \frac{2\pi r}{v} \)
\( T = \frac{2 \times \pi \times 7.82 \times 10^6}{7.13 \times 10^3} \)

Step5: Compute orbital period

First calculate numerator: \( 2 \times \pi \times 7.82 \times 10^6 \approx 4.91 \times 10^7 \)
Divide by \( v \): \( T \approx \frac{4.91 \times 10^7}{7.13 \times 10^3} \approx 6.89 \times 10^3\ \text{seconds} \)
(Convert to hours: \( \frac{6.89 \times 10^3}{3600} \approx 1.91\ \text{hours} \))

Step1: Orbital speed of Moon setup

Use \( v = \frac{2\pi r}{T} \), where \( r=3.84 \times 10^8\ \text{m} \), \( T=2358720\ \text{s} \)
\( v = \frac{2 \times \pi \times 3.84 \times 10^8}{2358720} \)

Step2: Compute orbital speed

Calculate numerator: \( 2 \times \pi \times 3.84 \times 10^8 \approx 2.41 \times 10^9 \)
Divide by \( T \): \( v \approx \frac{2.41 \times 10^9}{2358720} \approx 1.02 \times 10^3\ \text{m/s} \)

Step3: Centripetal acceleration setup

Use \( a_c = \frac{v^2}{r} \)
\( a_c = \frac{(1.02 \times 10^3)^2}{3.84 \times 10^8} \)

Step4: Compute centripetal acceleration

Calculate numerator: \( (1.02 \times 10^3)^2 = 1.04 \times 10^6 \)
Divide by \( r \): \( a_c \approx \frac{1.04 \times 10^6}{3.84 \times 10^8} \approx 2.71 \times 10^{-3}\ \text{m/s}^2 \)

Answer:

a. \( 7.13 \times 10^3\ \text{m/s} \)
b. \( 6.89 \times 10^3\ \text{seconds (or 1.91 hours)} \)

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Problem 9