Set the correct answer.
[tex]
\begin{array}{l}
G=6.67 \times 10^{-11} N m^2 / kg^2 \
M_e=5.598 \times 10^{24} kg \
v=\sqrt{\frac{G K}{r}}
\end{array}
[/tex]
What is the velocity of an Earth satellite that is in a circular orbit with a radius of [tex]$7.5 \times 10^7$[/tex] meters measured from the center of the Earth?
a. [tex]$29 \times 10^3$[/tex] meters
b. [tex]$23 \times 10^{-3}$[/tex] meters
c. [tex]$75 \times 10^3$[/tex] meters
d. [tex]$50-\omega^3$[/tex].
Answer (2)
Substitute the given values into the formula v = r G M e .
Calculate the value inside the square root: 7.5 × 1 0 7 ( 6.67 × 1 0 − 11 ) ( 5.598 × 1 0 24 ) = 4.978488 × 1 0 6 .
Take the square root: v = 4.978488 × 1 0 6 ≈ 2231.25 m / s .
The calculated velocity is approximately 2231.25 m / s , which does not match any of the options exactly. There might be a typo in the options. 2231.25 m / s
Explanation
Understanding the Problem We are given the gravitational constant G = 6.67 \t × 1 0 − 11 N m 2 / k g 2 , the mass of the Earth M e = 5.598 \t × 1 0 24 k g , and the radius of the circular orbit r = 7.5 \t × 1 0 7 meters. We are also given the formula for the velocity of a satellite in a circular orbit: v = \t r G M e . Our goal is to calculate the velocity v using the given values.
Substituting the Values Now, we substitute the given values into the formula: v = 7.5 × 1 0 7 m ( 6.67 × 1 0 − 11 N m 2 / k g 2 ) ( 5.598 × 1 0 24 k g )
Calculating the Value Inside the Square Root Let's calculate the value inside the square root: 7.5 × 1 0 7 ( 6.67 × 1 0 − 11 ) ( 5.598 × 1 0 24 ) = 7.5 × 1 0 7 3.733866 × 1 0 14 = 4.978488 × 1 0 6 So, we have: v = 4.978488 × 1 0 6
Taking the Square Root Now, we take the square root: v = 4.978488 × 1 0 6 ≈ 2231.25 m / s
Comparing with the Options Now we need to compare this result with the given options. The options are: a. 29 × 1 0 3 meters b. 23 × 1 0 − 3 meters c. 75 × 1 0 3 meters d. 50 − ω 3
Our calculated velocity is approximately 2231.25 m / s . We can rewrite this as 2.23125 × 1 0 3 m / s . None of the options match exactly. However, if we look at the order of magnitude, we can see that the closest option is 2.23125 × 1 0 3 m / s which is approximately equal to 2.3 × 1 0 3 m / s . However, there seems to be a typo in option B. It should probably be 2.3 × 1 0 3 instead of 23 × 1 0 − 3 .
However, since we don't have this option, let's re-evaluate the calculation. We have v = r G M e = 7.5 × 1 0 7 6.67 × 1 0 − 11 × 5.598 × 1 0 24 = 2231.25 m / s
Converting to km / s , we get 2.23 km / s . The closest answer in the options is A, which is 29 × 1 0 3 m / s = 29 km / s . There seems to be an error in the problem statement or the options provided. However, based on the calculation, the velocity is approximately 2.23 × 1 0 3 m / s .
Since none of the options are close to the calculated value, it is possible that there is a typo in the question or the options. However, based on the calculation, the velocity is approximately 2231.25 m/s, which is closest to 2.23 x 10^3 m/s. If we assume option A is 2.23 x 10^3 m/s, then option A would be the closest answer.
Examples
Calculating the orbital velocity of satellites is crucial in space mission planning. For example, when launching a communication satellite into a geostationary orbit, engineers need to precisely determine the required velocity to ensure the satellite stays in its designated position above Earth. Similarly, understanding orbital velocity is essential for missions involving the International Space Station (ISS) or other Earth-orbiting spacecraft, ensuring they maintain stable orbits and can perform their intended functions effectively. This calculation helps in designing efficient trajectories and managing fuel consumption for space vehicles.
The velocity of the Earth satellite in a circular orbit with a radius of 7.5 × 1 0 7 meters is approximately 2231.25 m/s. This calculated value does not match the provided answer options, suggesting there may be a typographical error. Therefore, none of the options accurately reflect the calculated orbital velocity.
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