Question
Let G be the universal gravitational constant and mp be the mass of the planet a satellite is orbiting. Which equation could be used to find the velocity of the satellite if it is placed in a low Earth orbit?(1 point)
Responses
v=Gmp(7,000 km)−−−−−−−√
v is equal to square root of the fraction with numerator cap g m sub p and denominator left bracket 7 comma 000 kilometers right bracket end root
v=1(200 km)Gmp−−−−√
v is equal to the fraction with numerator 1 and denominator right paran 200 kilometers left paran end fraction square root of cap g m sub p end root
v=Gmp(200 km)−−−−−−√
v is equal to square root of the fraction with numerator cap g m sub p and denominator right paran 200 kilometers left paran end root
v=1(7,000 km)Gmp−−−−√
v is equal to the fraction with numerator 1 and denominator left bracket 7 comma 000 kilometers right bracket end fraction square root of cap g m sub p end root
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Responses
v=Gmp(7,000 km)−−−−−−−√
v is equal to square root of the fraction with numerator cap g m sub p and denominator left bracket 7 comma 000 kilometers right bracket end root
v=1(200 km)Gmp−−−−√
v is equal to the fraction with numerator 1 and denominator right paran 200 kilometers left paran end fraction square root of cap g m sub p end root
v=Gmp(200 km)−−−−−−√
v is equal to square root of the fraction with numerator cap g m sub p and denominator right paran 200 kilometers left paran end root
v=1(7,000 km)Gmp−−−−√
v is equal to the fraction with numerator 1 and denominator left bracket 7 comma 000 kilometers right bracket end fraction square root of cap g m sub p end root
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Answers
GPT-4o mini
To find the velocity of a satellite in a low Earth orbit, we can use the formula for the orbital velocity (\( v \)) of a satellite, which is given by:
\[
v = \sqrt{\frac{G m_p}{r}}
\]
where:
- \( G \) is the universal gravitational constant,
- \( m_p \) is the mass of the planet (Earth),
- \( r \) is the distance from the center of the planet to the satellite.
In a low Earth orbit, the altitude is typically about 200 km to 700 km above the surface of the Earth. However, the value of \( r \) will be the sum of the Earth's radius (approximately 6,371 km) and the altitude of the satellite.
For a low Earth orbit of approximately 200 km, the distance \( r \) would be approximately:
\[
r \approx 6,371 \text{ km} + 200 \text{ km} = 6,571 \text{ km}
\]
For an orbit at about 7,000 km from the center of the Earth, we can use \( r \approx 7,000 \text{ km} \).
From the available options provided, the closest and most accurate representation of the equation for the velocity of a satellite, given the stated parameters, is:
\[
v = \sqrt{\frac{G m_p}{7,000 \text{ km}}}
\]
This indicates that the answer is:
**v = Gmp(7,000 km)−−−−−−−√** (using the notation provided).
Please note that since the correct format might not match the typical mathematical expressions, ensure that your understanding of the parameters and units is correct based on the context of the question.
\[
v = \sqrt{\frac{G m_p}{r}}
\]
where:
- \( G \) is the universal gravitational constant,
- \( m_p \) is the mass of the planet (Earth),
- \( r \) is the distance from the center of the planet to the satellite.
In a low Earth orbit, the altitude is typically about 200 km to 700 km above the surface of the Earth. However, the value of \( r \) will be the sum of the Earth's radius (approximately 6,371 km) and the altitude of the satellite.
For a low Earth orbit of approximately 200 km, the distance \( r \) would be approximately:
\[
r \approx 6,371 \text{ km} + 200 \text{ km} = 6,571 \text{ km}
\]
For an orbit at about 7,000 km from the center of the Earth, we can use \( r \approx 7,000 \text{ km} \).
From the available options provided, the closest and most accurate representation of the equation for the velocity of a satellite, given the stated parameters, is:
\[
v = \sqrt{\frac{G m_p}{7,000 \text{ km}}}
\]
This indicates that the answer is:
**v = Gmp(7,000 km)−−−−−−−√** (using the notation provided).
Please note that since the correct format might not match the typical mathematical expressions, ensure that your understanding of the parameters and units is correct based on the context of the question.