Answer:
A. 5521.86 secs
B. 7680.65 m/s
Explanation:
Parameters given:
Radius of orbit of the space station, R = 6380000 + 370000 = 6750000 m
Mass of earth = 5.97 * 10^24 kg
Gravitational constant, G = 6.67 * 10^(-11) Nm²/kg²
A. Orbital period, T, can be obtained using the formula:
T²/R³ = (4 * pi²) / (G * M)
T²/(6750000³) = (4 * 3.142²) / (6.67 * 10^(-11) * 5.97 * 10^24)
T² = (4 * 3.142² * 6750000³) / (6.67 * 10^(-11) * 5.97 * 10^24)
T² = 30490944.39
T = 5521.86 secs
B. Orbital velocity, v, can be obtained by using the formula:
v² = (G * M) / R
v² = (6.67 * 10^(-11) * 5.97 * 10^(-24)) / 6750000
v² = 58992384
v = 7680.65 m/s
Lineart
Answer: the answer is lineart
If time (t), acceleration(a) and initial velocity (Vi) is given distance can be find by
S = (Vi)t + (1/2) at²
If both velocities initial velocity an final velocity is given then formula will be
2aS = (Vf)²- (Vi²)
Where S is the distance.
Answer:
G = 1.670 ×
Explanation:
From Newton's law of universal gravitation,
F =
Where F is the force of attraction, G is the gravitational constant, M is the mass of the earth, R is the radius of the earth.
From Newton's second law of motion,
F = mg
mg =
g =
⇒ G =
If the radius of the earth becomes half,
R = , then;
G =
Given that: g = 9.8 m/, radius of the earth is 6371000 m, mass of the earth is 5.972 × kg, then;
G =
= 1.670 ×
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