An artificial satellite circles the Earth in a circular orbit at a location where the acceleration due to gravity is 8.26 m/s2. Determine the orbital period of the satellite.
To determine the orbital period of the satellite, we can use the formula for the period of a satellite in a circular orbit:
T = 2π√(r³/g)
where:
T = orbital period
π = 3.14159 (constant)
r = radius of the satellite's orbit
g = acceleration due to gravity
In this case, we are given that the acceleration due to gravity is 8.26 m/s². To find the radius of the satellite's orbit, we need to use the value of the acceleration due to gravity and the formula for acceleration due to gravity at a given height:
g = GM/r²
where:
G = gravitational constant (approximately 6.674 × 10⁻¹¹ N(m/kg)²)
M = mass of the Earth (approximately 5.972 × 10²⁴ kg)
r = radius of the satellite's orbit
We can rearrange this equation to solve for r:
r² = GM/g
Substituting the given values, we have:
r² = (6.674 × 10⁻¹¹ N(m/kg)²)(5.972 × 10²⁴ kg)/(8.26 m/s²)
Simplifying:
r² = (6.674 × 5.972) × 10⁻¹¹ × 10²⁴ / 8.26
r² ≈ 3.970 × 10⁶ × 10⁶ / 8.26
r² ≈ 4.81067 × 10¹²
Taking the square root:
r ≈ √(4.81067 × 10¹²)
r ≈ 6.936 × 10⁶ meters
Now, substituting the value of r into the formula for the orbital period:
T = 2π√(r³/g)
T = 2π√((6.936 × 10⁶)³/8.26)
Evaluating this expression:
T = 2π√(2.4420846 × 10²⁰ / 8.26)
T ≈ 2π√(2.95677 × 10¹⁸)
T ≈ 2π× 5.4364 × 10⁹
T ≈ 34.169 × 10⁹ seconds
Converting to hours:
T ≈ (34.169 × 10⁹) / 3600
T ≈ 9.491 × 10⁶ hours
Therefore, the orbital period of the satellite is approximately 9.491 × 10⁶ hours.
To determine the orbital period of a satellite, we can use Kepler's Third Law of Planetary Motion, which states that the square of the orbital period is proportional to the cube of the semi-major axis of the orbit.
First, we need to find the value of the semi-major axis, which is the radius of the circular orbit of the satellite. To do this, we can use the acceleration due to gravity, which is given as 8.26 m/s^2.
The acceleration due to gravity on an object in circular orbit can be calculated using the formula:
a = (GM)/r^2,
where a is the acceleration due to gravity, G is the gravitational constant (approximately 6.67430 x 10^-11 Nm^2/kg^2), M is the mass of the Earth (approximately 5.972 x 10^24 kg), and r is the radius of the orbit.
Rearranging the formula, we can solve for r:
r = sqrt(GM/a).
Plugging in the known values, we have:
r = sqrt((6.67430 x 10^-11 Nm^2/kg^2 * 5.972 x 10^24 kg) / 8.26 m/s^2).
Now, we can calculate the semi-major axis, which is simply the radius:
a = r.
Finally, we can use the formula for the orbital period, T:
T = 2π sqrt(a^3 / GM).
Substituting the values we have calculated, we can find the orbital period of the satellite.
v^2/r = 8.26
F/m = local g = G M/r^2 = 8.26
6.67*10^-11 * Mass earth/r^2 = 8.26
solve that for r
then
v^2 = 8.26 r
solve for v
v T = 2 pi r
T = 2 pi r/v