a sitealite is orbiting Earth at an orbital radius of 3.9×103 m. What is its velocity in terms of G and the mass of Earth, mp? (Ignore units in the answer.)(1 point)
Responses
(2.6×10−4)Gmp−−−−√
(6.2×101)Gmp−−−−√
(1.6×10−2)Gmp−−−−√
(3.9×103)Gmp−−−−√
The velocity of the satellite in terms of G and the mass of Earth, mp, can be calculated using the formula for the orbital velocity:
v = √(G * mp / r)
where r is the orbital radius.
Plugging in the values:
v = √(G * mp / 3.9×10^3)
Therefore, the correct answer is (3.9×103)Gmp−−−−√
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 geostationary orbit?(1 point)
Responses
v=Gmp(15,522 km)−−−−−−−−√
v=Gmp(48,115 km)−−−−−−−−√
v=Gmp(7,324 km)−−−−−−−√
v=Gmp(42,164 km)−−−−−−−−√
To find the velocity of a satellite in a geostationary orbit, we can use the equation:
v = √(G * mp / r)
where r is the radius of the orbit.
In a geostationary orbit, the satellite remains fixed above a specific point on the Earth's surface. The radius of this orbit is equal to the radius of the Earth itself. The value is approximately 6,371 km (or 6,371,000 m).
Therefore, the correct equation would be:
v = √(G * mp / 6,371,000)
Hence, none of the given options are correct.
To find the velocity of the satellite in terms of G (the gravitational constant) and the mass of Earth (mp), we can use the formula for the velocity of a satellite in circular orbit:
v = √(G * mp / r)
Where:
v = velocity of the satellite
G = gravitational constant
mp = mass of Earth
r = orbital radius of the satellite
In this case, the orbital radius is given as 3.9x10^3 m. Plugging in these values into the formula, we get:
v = √(G * mp / 3.9x10^3)
Therefore, the correct answer is:
(3.9x10^3)Gmp−−−−√
To find the velocity of a satellite orbiting Earth at a given orbital radius, we can use the formula for orbital velocity:
v = √(G * M / r)
Where:
v = velocity of the satellite
G = gravitational constant (approximately 6.67 x 10^-11 m^3 kg^-1 s^-2)
M = mass of Earth
r = orbital radius
In this case, the orbital radius is given as 3.9×10^3 m.
To determine the correct answer choice, we can substitute the values into the formula and solve for the velocity. Let's go through each answer choice:
(2.6×10−4)Gmp−−−−√:
Substituting the values into the formula:
v = √((2.6×10−4)Gmp / (3.9×10^3))
This does not match the formula, as the square root is applied to only part of the expression.
(6.2×101)Gmp−−−−√:
Substituting the values into the formula:
v = √((6.2×10^1)Gmp / (3.9×10^3))
This does not match the formula, as the square root is applied to only part of the expression.
(1.6×10−2)Gmp−−−−√:
Substituting the values into the formula:
v = √((1.6×10−2)Gmp / (3.9×10^3))
This does not match the formula, as the square root is applied to only part of the expression.
(3.9×10^3)Gmp−−−−√:
Substituting the values into the formula:
v = √((3.9×10^3)Gmp / (3.9×10^3))
Simplifying the expression:
v = √(Gmp)
This matches the formula for orbital velocity and the square root is applied correctly. Therefore, the correct answer choice is (3.9×10^3)Gmp−−−−√.