Suppose you are at mission control on the moon, in charge of launching a moon-orbiting communications satellite.
Moon mass =7.36×10 22
kg Moon radius =1.74×10 6
m
a. First, how much would a
1500 kg
satellite weigh near the surface of the moon? b. The satellite is to have an altitude of
100 km
above the moon's surface. What is the radius of the orbit of the satellite? c. When the satellite is in orbit, how big will the centripetal force be? Explain. d. Find the required orbital velocity for the satellite. e. How long will it take the satellite to orbit the moon? (This time is called the orbital period.) f. Is this satellite accelerating while in orbit? If so, what is the direction and magnitude of the acceleration?

a. The weight of the satellite near the surface of the moon would be the mass of the satellite multiplied by the acceleration due to gravity on the moon. The acceleration due to gravity on the moon is about 1.62 m/s^2, so the weight of the satellite would be

1500 kg * 1.62 m/s^2 = 2430 N.

b. The radius of the orbit of the satellite would be the distance from the center of the moon to the altitude of the satellite. The radius of the moon is 1.74 x 10^6 m, so the radius of the orbit of the satellite would be

1.74 x 10^6 m + 100 km = 1.84 x 10^6 m.

c. The centripetal force is the force that is required to keep an object in circular motion. It is equal to the mass of the object times its velocity squared, divided by the radius of its orbit. To find the centripetal force on the satellite, we need to know its velocity. We can find this using the equation for the centripetal force, which is:

F = m * v^2 / r

where F is the centripetal force, m is the mass of the object, v is the velocity of the object, and r is the radius of the orbit.

d. To find the required orbital velocity for the satellite, we can rearrange the equation for the centripetal force to solve for v:

v = sqrt(F * r / m)

Plugging in the values, we get:

v = sqrt(2430 N * 1.84 x 10^6 m / 1500 kg) = 2157 m/s

e. The orbital period is the time it takes for an object to complete one orbit. It is equal to the circumference of the orbit divided by the velocity of the object. The circumference of the orbit is 2 * pi * r, where r is the radius of the orbit. The velocity of the object is the orbital velocity we calculated above. Plugging these values into the equation, we get:

T = 2 * pi * r / v

Plugging in the values, we get:

T = 2 * pi * 1.84 x 10^6 m / 2157 m/s = 2.87 x 10^4 s

f. Yes, the satellite is accelerating while in orbit. The direction of the acceleration is towards the center of the orbit, and the magnitude of the acceleration is the centripetal acceleration, which can be calculated using the equation:

a = v^2 / r

Plugging in the values, we get:

a = 2157 m/s^2 / 1.84 x 10^6 m = 1.17 x 10^-3 m/s^2