Question

1. how high above earth’s surface must a satellite be if it has a speed of 6600 m/s?
2. if a moons orbit period is 2.45×10⁶ s and it is 3.75×10⁸ m away, what is the speed of the moon in its orbit around the planet and the mass of the planet?

if a small planet were located 8 times as far as the earth’s distance from the sun, what would be its period in earth years?
for the enterprise to use its transporter it must be in synchronous orbit over
that height above the planet mars must the enterprise be for a synchronous

Explanation:

Problem 2:

Step1: Recall orbital speed formula

The orbital speed \( v \) of a satellite is given by \( v = \sqrt{\frac{GM}{r}} \), where \( G = 6.67\times 10^{-11}\ \text{N m}^2/\text{kg}^2 \), \( M = 5.97\times 10^{24}\ \text{kg} \) (mass of Earth), and \( r \) is the distance from the center of Earth to the satellite. We can rearrange for \( r \): \( r=\frac{GM}{v^{2}} \)

Step2: Calculate \( r \)

Substitute \( G = 6.67\times 10^{-11}\), \( M = 5.97\times 10^{24}\), and \( v = 6600\ \text{m/s} \) into the formula:

$$ r=\frac{6.67\times 10^{-11}\times5.97\times 10^{24}}{(6600)^{2}} $$

First, calculate the numerator: \( 6.67\times 10^{-11}\times5.97\times 10^{24}=6.67\times5.97\times 10^{13}\approx 39.82\times 10^{13}=3.982\times 10^{14} \)
Denominator: \( (6600)^{2}=43560000 = 4.356\times 10^{7} \)
Then \( r=\frac{3.982\times 10^{14}}{4.356\times 10^{7}}\approx 9.14\times 10^{6}\ \text{m} \)

Step3: Find height above Earth's surface

The radius of Earth \( R = 6.37\times 10^{6}\ \text{m} \). The height \( h=r - R \)
\( h=9.14\times 10^{6}-6.37\times 10^{6}=2.77\times 10^{6}\ \text{m} \) (or 2770 km)

Step1: Recall orbital speed formula

The orbital speed \( v \) is given by \( v=\frac{2\pi r}{T} \), where \( r = 3.75\times 10^{8}\ \text{m} \) and \( T = 2.45\times 10^{6}\ \text{s} \)

Step2: Calculate \( v \)

Substitute the values:

$$ v=\frac{2\pi\times3.75\times 10^{8}}{2.45\times 10^{6}} $$

First, calculate the numerator: \( 2\pi\times3.75\times 10^{8}\approx 2.356\times 10^{9} \)
Then divide by denominator: \( v=\frac{2.356\times 10^{9}}{2.45\times 10^{6}}\approx 961.6\ \text{m/s} \)

Part 2: Mass of the planet

Step1: Recall Kepler's third law (or gravitational force formula)

From \( \frac{GMm}{r^{2}}=m\frac{4\pi^{2}r}{T^{2}} \), we can solve for \( M \): \( M=\frac{4\pi^{2}r^{3}}{GT^{2}} \)

Step2: Substitute values

\( G = 6.67\times 10^{-11}\), \( r = 3.75\times 10^{8}\), \( T = 2.45\times 10^{6} \)
First, calculate \( r^{3}=(3.75\times 10^{8})^{3}=3.75^{3}\times 10^{24}\approx 52.73\times 10^{24}=5.273\times 10^{25} \)
\( T^{2}=(2.45\times 10^{6})^{2}=6.00\times 10^{12} \) (approx)
Then \( M=\frac{4\pi^{2}\times5.273\times 10^{25}}{6.67\times 10^{-11}\times6.00\times 10^{12}} \)
Calculate numerator: \( 4\pi^{2}\times5.273\times 10^{25}\approx 4\times9.87\times5.273\times 10^{25}\approx 207.7\times 10^{25}=2.077\times 10^{27} \)
Denominator: \( 6.67\times 6.00\times 10^{1}\approx 40.02\times 10^{1}=4.002\times 10^{2} \)
\( M=\frac{2.077\times 10^{27}}{4.002\times 10^{2}}\approx 5.19\times 10^{24}\ \text{kg} \) (similar to Earth's mass, which makes sense for a planet with a moon)

Answer:

The satellite must be approximately \( 2.77\times 10^{6}\ \text{m} \) (or 2770 km) above Earth's surface.

Problem 3:

Part 1: Speed of the moon