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