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In order to better map the surface features of the Moon, a 361 kg imaging satellite is put into circular orbit around the Moon at an altitude of 147 km. Calculate the satellite's kinetic energy \( K \), gravitational potential energy \( U \), and total orbital energy \( E \).

The radius and mass of the Moon are 1740 km and \( 7.36 \times 10^{22} \) kg.

Answer :

Answer:

Explanation:

Mass of satellite

M_s = 361 kg

Distance of satellite from moon

h = 147 km = 147,000m

Radius of the moon is

R_m = 1740 km = 1740,000m

Mass of the moon is

M_m = 7.36 × 10²² kg.

The kinetic energy is equal to the potential energy of the body to the surface of the moon from the conservation of energy.

K.E = P.E = mgh

Gravity on moon is g = 1.62 m/s²

K.E = 361 × 1.62 × 147,000

K.E = 8.597 × 10^7 J.

B. The gravitational potential energy can be calculated using

U = G•M_s × M_m (1/R_s - 1 / R)

R is the total distance from the centre of the moon to the satellite

R = h + R_m = 147 + 1740 = 1887km

R = 1,887,000 m

U = 6.67 × 10^-11 × 361 × 7.36 × 10²² (1/1,740,000 - 1/1,887,000)

U = 6.67 × 10^-11 × 361 × 7.36 × 10²² × 4.48 × 10^-8

U = 7.93 × 10^7 J

Then,

The total energy becomes

E = K.E + U

E= 8.597 × 10^7 + 7.93 × 10^7 J

E = 1.653 × 10^8 J

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