A satellite of mass 205 kg is launched from a site on Earth's equator into an orbit at 200 km above the surface of Earth.
(a) Assuming a circular orbit, what is the orbital period of this satellite?
s
(b) What is the satellite's speed in its orbit?
m/s
(c) What is the minimum energy necessary to place the satellite in orbit, assuming no air friction?
J
To find the orbital period of the satellite, we can use Kepler's third law, which states that the square of the orbital period is proportional to the cube of the semi-major axis of the orbit. Let's assume the radius of the Earth is 6,371 km.
(a) First, let's find the semi-major axis of the satellite's circular orbit:
The altitude of the satellite above the surface of the Earth is 200 km.
The total distance from the center of the Earth to the satellite's orbit is the sum of the radius of the Earth and the altitude of the satellite:
Total distance = Earth's radius + Satellite's altitude
= 6,371 km + 200 km
= 6,571 km
= 6,571,000 m
The semi-major axis is half of the total distance:
Semi-major axis = Total distance / 2
= 6,571,000 m / 2
= 3,285,500 m
Now, let's plug the values into Kepler's third law to find the orbital period:
T² = (4π² / G) * (a³ / M)
Where:
T = Orbital period (unknown)
G = Gravitational constant (6.67430 × 10⁻¹¹ m³/kg/s²)
a = Semi-major axis (3,285,500 m)
M = Mass of the Earth (5.972 × 10²⁴ kg)
T² = (4π² / 6.67430 × 10⁻¹¹) * (3,285,500)³ / (5.972 × 10²⁴)
T² = 4π² * (3,285,500)³ / (6.67430 × 10⁻¹¹ * 5.972 × 10²⁴)
T² = (4 * (3.14159265359)² * (3,285,500)³) / (6.67430 × 10⁻¹¹ * 5.972 × 10²⁴)
T² ≈ 1.095403249
Taking the square root of both sides:
T ≈ √1.095403249
T ≈ 1.046041117
Therefore, the orbital period of the satellite is approximately 1.046 seconds.
(b) To find the satellite's speed in its orbit, we can use the formula for the speed of an object in circular motion:
v = 2πr / T
Where:
v = Speed (unknown)
π = Pi (approximately 3.14159)
r = Radius of the orbit (Earth's radius + Altitude of the satellite)
T = Orbital period (1.046 seconds)
r = 6,371 km + 200 km
r = 6,571 km
r = 6,571,000 m
v = 2π * 6,571,000 m / 1.046 s
v = 2π * 6,571,000 m / 1.046 s
v ≈ 38,969,700 m/s
Therefore, the satellite's speed in its orbit is approximately 38,969,700 m/s.
(c) The minimum energy necessary to place the satellite in orbit assuming no air friction is the sum of its kinetic energy and gravitational potential energy.
For the kinetic energy, we use the formula:
KE = 1/2 * M * v²
Where:
KE = Kinetic Energy (unknown)
M = Mass of the satellite (205 kg)
v = Speed of the satellite (38,969,700 m/s)
KE = 1/2 * 205 kg * (38,969,700 m/s)²
KE ≈ 1/2 * 205 kg * (1.5192 × 10¹⁵ m²/s²)
KE ≈ 7.83568 × 10¹⁶ J
For the gravitational potential energy, we use the formula:
PE = -G * M * m / r
Where:
PE = Gravitational Potential Energy (unknown)
G = Gravitational constant (6.67430 × 10⁻¹¹ m³/kg/s²)
M = Mass of the Earth (5.972 × 10²⁴ kg)
m = Mass of the satellite (205 kg)
r = Radius of the orbit (6,571,000 m)
PE = -6.67430 × 10⁻¹¹ * 5.972 × 10²⁴ kg * 205 kg / 6,571,000 m
PE ≈ -1.107 × 10³ J
The minimum energy necessary to place the satellite in orbit is the sum of the kinetic energy and gravitational potential energy:
Total Energy = KE + PE
Total Energy ≈ 7.83568 × 10¹⁶ J + (-1.107 × 10³ J)
Total Energy ≈ 7.83568 × 10¹⁶ J
Therefore, the minimum energy necessary to place the satellite in orbit, assuming no air friction, is approximately 7.83568 × 10¹⁶ J.
To find the orbital period of the satellite, we can use the formula:
T = 2π √(R³/GMe)
Where:
T = Orbital period
π = Pi (approximately 3.14159)
R = Distance from the center of the Earth to the satellite's orbit (radius of Earth + altitude of the orbit)
G = Gravitational constant (approximately 6.67430 x 10^-11 m³/kg/s²)
Me = Mass of the Earth (approximately 5.97219 x 10^24 kg)
First, we need to calculate the distance, R, from the center of the Earth to the satellite's orbit:
Radius of Earth (R_e) = 6,371 km
Altitude of orbit (h) = 200 km
R = R_e + h = 6,371 km + 200 km = 6,571 km = 6,571,000 m
Now we can substitute the values into the formula for T:
T = 2π √((6,571,000 m)³ / (6.67430 x 10^-11 m³/kg/s² * 5.97219 x 10^24 kg))
Simplifying further:
T = 2π √(2.44962 x 10^21 m³ / 39.8215 m³/s²)
T = 2π √(6.15531 x 10^18)
T ≈ 2π * 7.84303 x 10^9 seconds
T ≈ 49,213,633 seconds
So the orbital period of the satellite is approximately 49,213,633 seconds.
To determine the satellite's speed in its orbit, we can use the formula:
v = 2πR / T
Where:
v = Speed
R = Distance from the center of the Earth to the satellite's orbit (6,571,000 m, as calculated earlier)
T = Orbital period (49,213,633 seconds, as calculated earlier)
Substituting the values into the formula:
v = 2π(6,571,000 m) / 49,213,633 seconds
v ≈ 1672.07 m/s
So the satellite's speed in its orbit is approximately 1672.07 m/s.
To find the minimum energy necessary to place the satellite in orbit, we can use the formula for gravitational potential energy:
E = -GMeMs / 2R
Where:
E = Gravitational potential energy
G = Gravitational constant (approximately 6.67430 x 10^-11 m³/kg/s²)
Me = Mass of the Earth (approximately 5.97219 x 10^24 kg)
Ms = Mass of the satellite (205 kg)
R = Distance from the center of the Earth to the satellite's orbit (6,571,000 m, as calculated earlier)
Substituting the values into the formula:
E = -(6.67430 x 10^-11 m³/kg/s² * 5.97219 x 10^24 kg * 205 kg) / 2 * 6,571,000 m
E ≈ -1.763 x 10^11 J
The minimum energy necessary to place the satellite in orbit, assuming no air friction, is approximately -1.763 x 10^11 Joules. Note that the negative sign indicates that the satellite has negative potential energy in orbit.
200*10^3 = .2*10^6 m high
R earth = 6.38*10^6
so R = 6.58 * 10^6 meters
G = 6.67384 × 10-11 m3 kg-1 s-2
earth mass = Me = 5.97*10^24 kg
m v^2/R = G Me m /R^2
v^2 = G Me/R
= 6.67 * 5.97 /6.58 * 10^(-11+24-6)
= 6.05 * 10^7 = 60.5 * 10^6
so
v = 7.78*10^3 = 7,780 m/s (PART B)
C = 2 pi R = 2 pi *6.58*10^6
= 41.3 * 10^6 meters circumference
so T = 41.3*10^6/7.78*10^3
= 5.31 * 10^3 seconds (PART A) = 1.47 hours
Work done to increase r from Re to R + Ke
potential energy = U call it 0 at infinity
U = -G Me m/r
U at R = -G Me m/R
U at Re = -G Me m /Re
U at R - U at Re = G Me m /Re-G Me m/R
= G Me m(R-Re)/(R Re)
then Ke = (1/2)m v^2
= (1/2) m G Me/R
sum of U + Ke =
G Me m(R-Re)/(R Re) + (1/2) G Me m Re/(R Re)
= (G Me m )(R-Re+.5Re)/(R Re)
= (G Me m)(R-.5Re)/(R Re)
since R and Re are about the same really
= about (1/2)G Me m /R