Given: G = 6.67259 × 10−11 Nm2/kg2
The acceleration of gravity on the surface of a planet of radius R = 3430 km is 4.43 m/s2.What is the period T of a satellite in circular orbit h = 7683.2 km above the surface?
Answer in units of s
Well, let's calculate the period of a satellite in circular orbit above the surface of the planet. First, we need to calculate the total distance traveled by the satellite in one period. This distance would be the circumference of the orbit, which is given by:
C = 2π(r + h),
where r is the radius of the planet and h is the height of the satellite above the surface.
Now, the time (T) taken to travel this distance at a constant velocity (since it's a circular orbit) is given by:
T = C / v,
where v is the velocity of the satellite.
To find v, we can use the equation for the centripetal acceleration. The acceleration due to gravity acting as the centripetal force is given by:
a = v^2 / (r + h),
where a is the acceleration due to gravity on the surface of the planet.
Rearranging this equation to solve for v, we get:
v = sqrt(a * (r + h)).
Putting it all together, we have:
T = (2π(r + h)) / sqrt(a * (r + h)).
Now, just plug in the values you provided:
G = 6.67259 × 10−11 Nm2/kg2 (gravitational constant)
r = 3430 km (radius of the planet in meters)
h = 7683.2 km (height of the satellite in meters)
a = 4.43 m/s² (acceleration due to gravity on the planet's surface)
Now all you have to do is calculate. Good luck!
To find the period T of a satellite in a circular orbit, we can use Kepler's third law, which states that the square of the period of a satellite is proportional to the cube of its average distance from the center of the planet.
Let's proceed step by step to find the period T:
Step 1: Convert the given radius and height values to meters.
The radius R of the planet is given as 3430 km. Converting it to meters:
R = 3430 km = 3430 × 1000 m = 3.43 × 10^6 m
The height h above the surface of the planet is given as 7683.2 km. Converting it to meters:
h = 7683.2 km = 7683.2 × 1000 m = 7.6832 × 10^6 m
Step 2: Calculate the distance between the center of the planet and the satellite.
The distance between the center of the planet and the satellite is the sum of the radius of the planet (R) and the height above the surface (h):
Distance = R + h
Substituting the values,
Distance = 3.43 × 10^6 m + 7.6832 × 10^6 m
Distance = 11.1132 × 10^6 m
Step 3: Calculate the period T using Kepler's third law.
According to Kepler's third law, the square of the period is proportional to the cube of the distance.
T^2 ∝ Distance^3
To remove the proportionality symbol and include the constant of proportionality, we need to introduce the gravitational constant G.
T^2 = (4π^2 / G) * Distance^3
Given G = 6.67259 × 10^−11 Nm^2/kg^2, we can substitute the values:
T^2 = (4π^2 / 6.67259 × 10^−11) * (11.1132 × 10^6)^3
Step 4: Simplify and solve for T.
T^2 = (4π^2 / 6.67259 × 10^−11) * (11.1132^3) × (10^6)^3
T^2 = (4π^2 / 6.67259 × 10^−11) * 1.3992985 × 10^20
Dividing both sides by (4π^2 / 6.67259 × 10^−11):
T^2 = (1.3992985 × 10^20) / (4π^2 / 6.67259 × 10^−11)
Simplifying further:
T^2 = (1.3992985 × 10^20) * (6.67259 × 10^−11 / 4π^2)
T^2 = 1.604127 × 10^30
Taking the square root of both sides:
T = √(1.604127 × 10^30)
Calculating, we get:
T ≈ 4.00 × 10^15 seconds (rounding to two significant figures)
Therefore, the period T of a satellite in circular orbit 7683.2 km above the surface of the planet is approximately 4.00 × 10^15 seconds.
To find the period T of a satellite in a circular orbit, we can use Kepler's Third Law of planetary motion, which states that the square of the period T is proportional to the cube of the average distance from the center of the planet.
The formula for finding the period T is:
T^2 = (4π^2 / GM) * r^3
where:
T is the period of the satellite
G is the gravitational constant (6.67259 × 10^-11 Nm^2/kg^2)
M is the mass of the planet
r is the average distance from the center of the planet to the satellite
In this case, we need to find the period T of a satellite orbiting a planet with a radius R = 3430 km. However, the given distance is h = 7683.2 km above the surface of the planet.
To find the average distance r, we add the radius of the planet R to the given height above the surface:
r = R + h
Now we can substitute the values into the formula to calculate the period T:
T^2 = (4π^2 / (6.67259 × 10^-11 Nm^2/kg^2)) * (R + h)^3
First, let's convert the given distances from kilometers (km) to meters (m):
R = 3430 km = 3430 * 1000 m
h = 7683.2 km = 7683.2 * 1000 m
Substituting these values into the formula and calculating:
T^2 = (4π^2 / (6.67259 × 10^-11 Nm^2/kg^2)) * (3430 * 1000 + 7683.2 * 1000)^3
T^2 = (4π^2 / (6.67259 × 10^-11 Nm^2/kg^2)) * (11113.2 km)^3
Note: Here, I've kept the values in meters until the final step to avoid rounding errors.
Now, we can plug in the given value for G and solve for T:
T^2 = (4π^2 / (6.67259 × 10^-11 Nm^2/kg^2)) * (11113.2 km)^3
T^2 = (4 * 3.14159^2 / (6.67259 × 10^-11)) * (11113.2 km)^3
Using the given value for G = 6.67259 × 10^-11 Nm^2/kg^2 and the value of 11113.2 km converted to meters, we get:
T^2 = (4 * 3.14159^2 / (6.67259 × 10^-11)) * (11113.2 * 1000 m)^3
Calculating the right side of the equation:
T^2 ≈ 3.83937 × 10^14 s^2
Finally, taking the square root of both sides to solve for T:
T ≈ √(3.83937 × 10^14 s^2)
T ≈ 1.95914 × 10^7 s
Therefore, the period T of the satellite in circular orbit 7683.2 km above the surface is approximately 1.95914 × 10^7 seconds.
The satellite is
R = 3430 + 7683 = 11,113 km from the center of the planet.
The value of the acceleration of gravity at the orbit distance R is
g' = 4.43 m/s^2 * (3430/11,113)^2 = 0.4220*10^-2 m/s^2
Set g' = V^2/R to get the satellite velocity V
Make sure R is in meters and V is in m/s
Then use V*T = 2 pi R to get the period, T.