NASA launches a satellite into orbit at a height above the surface of the Earth equal to the Earth's mean radius. The mass of the satellite is 550 kg. (Assume the Earth's mass is 5.97 1024 kg and its radius is 6.38 106 m.)
(a) How long does it take the satellite to go around the Earth once?
(b) What is the orbital speed of the satellite?
(c) How much gravitational force does the satellite experience?
To solve these problems, we can use the formulas for orbital period, orbital speed, and gravitational force.
(a) The formula for the period of an object in circular orbit is given by:
T = 2π√(R³/GM)
Where T is the period, R is the distance from the center of the Earth to the satellite, G is the gravitational constant (approximately 6.67430 x 10⁻¹¹ m³ kg⁻¹ s⁻²), and M is the mass of the Earth.
Given:
R = 6.38 x 10⁶ m
M = 5.97 x 10²⁴ kg
Substituting the given values into the formula:
T = 2π√((6.38 x 10⁶)³/(6.67430 x 10⁻¹¹)(5.97 x 10²⁴))
Calculating this expression will give us the time it takes for the satellite to go around the Earth once.
(b) The formula for the orbital speed of an object in circular orbit is given by:
v = √(GM/R)
Where v is the orbital speed.
Using the given values:
v = √((6.67430 x 10⁻¹¹)(5.97 x 10²⁴)/(6.38 x 10⁶))
Calculating this expression will give us the orbital speed of the satellite.
(c) The formula for the gravitational force experienced by a satellite in orbit is given by:
F = (GMm)/R²
Where F is the gravitational force, m is the mass of the satellite, G is the gravitational constant, M is the mass of the Earth, and R is the distance from the center of the Earth to the satellite.
Given:
m = 550 kg
G = 6.67430 x 10⁻¹¹ m³ kg⁻¹ s⁻²
M = 5.97 x 10²⁴ kg
R = 6.38 x 10⁶ m
Substituting the given values into the formula:
F = ((6.67430 x 10⁻¹¹)(5.97 x 10²⁴)(550))/(6.38 x 10⁶)²
Calculating this expression will give us the gravitational force experienced by the satellite.
To find the answers to these questions, we need to use some principles from physics, specifically the laws of motion and gravitation.
(a) To find the time it takes for the satellite to go around the Earth once, we can use the formula for the period of an object in circular motion:
T = 2πr / v
where T is the period, r is the radius, and v is the orbital speed.
In this case, since the satellite is launched at a height above the Earth's surface equal to the Earth's mean radius, the radius of the orbit is the sum of the Earth's radius and the satellite's height above the surface:
r = (6.38 x 10^6 m) + (6.38 x 10^6 m) = 12.76 x 10^6 m
Now, we need to find the orbital speed of the satellite first before we can calculate the period.
(b) The orbital speed of a satellite in a circular orbit is given by the formula:
v = √(GM / r)
where v is the orbital speed, G is the gravitational constant (approximately 6.67 x 10^-11 Nm^2/kg^2), M is the mass of the Earth, and r is the radius of the orbit.
Plugging in the values:
v = √((6.67 x 10^-11 Nm^2/kg^2)(5.97 x 10^24 kg) / (12.76 x 10^6 m))
Now, we can calculate the period using the formula mentioned earlier.
(c) The gravitational force experienced by the satellite can be found using the formula:
F = (GMm) / r^2
where F is the gravitational force, G is the gravitational constant, M is the mass of the Earth, m is the mass of the satellite, and r is the radius of the orbit.
Plugging in the values:
F = ((6.67 x 10^-11 Nm^2/kg^2)(5.97 x 10^24 kg)(550 kg)) / (12.76 x 10^6 m)^2
So, to summarize:
(a) To find the time it takes the satellite to go around the Earth once, use T = 2πr / v.
(b) To find the orbital speed of the satellite, use v = √(GM / r).
(c) To find the gravitational force experienced by the satellite, use F = (GMm) / r^2.
Plug in the given values into these formulas to find the answers.