Question

1) The moon has a mass of 7.4 × 1022 kg and completes an orbit of radius 3.8 × 108 m about every 28 days. The Earth has a mass of 6 × 1024 kg and completes an orbit of radius 1.5 × 1011 m every year. What is the speed of the Moon in its orbit? Answer in units of m/s.

2) What is the kinetic energy of the Moon in orbit?
Answer in units of J.

3. What is the kinetic energy of the Earth? Answer in units of J.

Answer 1

Mass of the moon Mm = 7.4 Ă— 10^22 kg
Moon orbital radius Rm= 3.8 Ă— 10^8 m
Time Tm = 28 days = 2.419 x 10^6
Mass of the Earth Me= 6 Ă— 10^24 kg
Earth orbital radius Re= 1.5 Ă— 10^11 m

Time Te = 365 days = 3.154 x 10^7
Length of Moon orbit Lm = 2 x pi x Rm = 2 x 3.14 x 3.8 Ă— 10^8 m = 23.86 x
10^8 m
Moon orbital velocity Vm = Lm / Tm = 23.86 x 10^8 m / 2.419 x 10^6 = 986.5 m/s
Kinetic Energy of the Moon = 1/2 x Mm x Vm^2 = 0.5 x 7.4 Ă— 10^22 x 986.5^2
Kinetic Energy of the Moon = 36 x 10^27 J
Length of earth orbit Le = 2 x pi x Re = 2 x 3.14 x 1.5 Ă— 10^11 = 9.42 x 10^11 m
Earth orbital Velocity Ve = Lm / Tm = 9.42 x 10^11 / 3.154 x 10^7 = 29866 m/s
Kinetic Energy of the Earth = 1/2 x Me x Ve^2 = 0.5 x 6 Ă— 10^24 x 29866^2
Kinetic Energy of the Earth = 26.76 x 10^32 J

Answer 2

What is the speed of the Moon in its orbit? Answer in units of m / s.
By definition we have that the speed in a circular movement is given by:
v = w * r
We then have to replace the values:
v = ((2 * pi) / (28 * (24/1) * (3600/1))) * (3.8 * 10 ^ 8)
v = 986.94 m / s

2) What is the kinetic energy of the Moon in orbit? Answer in units of J.

By definition, the kinetic energy is given by:
K = (1/2) * (m) * (V ^ 2)
Where,
m: mass
V: speed
Substituting the values ​​we have:
K = (1/2) * (7.4 * 10 ^ 22) * ((986.94) ^ 2)
K = 3.60E + 28 J

3. What is the kinetic energy of the Earth? Answer in units of J.
First we must find the speed of the earth:
v = ((2 * pi) / ((1) * (365/1) * (24/1) * (3600/1))) * (1.5 * 10 ^ 11)
v = 29886 m / s
By definition, the kinetic energy is given by:
K = (1/2) * (m) * (V ^ 2)
Where,
m: mass
V: speed
Substituting the values ​​we have:
K = (1/2) * (6 * 10 ^ 24) * ((29886) ^ 2)
K = 2.67E + 33 J