In order better to map the surface features of the moon, a 365 kg imaging satellite is put into circular orbit around the moon at an altitude of 105 km. calculate the satellite's kinetic energy, gravitational potential energy, and total orbital energy. the radius and mass of the moon are 1740 km and 7.36×1022 kg.

K = 4.85 x 10^8 J

U = - 9.71 x 10^8 J

T = - 4.85 x 10^8 J

Explanation:

mass of satellite, m = 365 kg

altitude, h = 105 km

radius of moon, R = 1740 km

mass of moon, M = 7.36 x 10^22 kg

r = R + h = 1740 + 105 = 1845 km

Kinetic energy of the satellite

K = G M m / 2r

K = (6.67 x 10^-11 x 7.36 x 10^22 x 365) / ( 2 x 1845 x 1000)

K = 4.85 x 10^8 J

Potential energy

U = - G M m / r

U = - (6.67 x 10^-11 x 7.36 x 10^22 x 365) / ( 1845 x 1000)

U = - 9.71 x 10^8 J

Total energy

T = U + K

T = (- 9.71 + 4.85) x 10^8

T = - 4.85 x 10^8 J

depends on what you mean by "group." for order, it would be coleoptera (beetles). for class, it would be insecta (insects). for phylum, it would be the arthropoda (animals with jointed exoskeletons).

explanation:

okay i have so many questions. but the big one here is: um wt..f is this mess? ?