Calculate the period of a satellite orbiting the Moon, 98 kmkm above the Moon's surface. Ignore effects of the Earth. The radius of the Moon is 1740 kmkm.

3.6*10^18s

Explanation:

To find the period of the satellite

We need to apply kephler's third law

Which is

MP² = (4π²/G) d³

d=semi-major axis which is the distance from center of moon = 98km+1740km = 1838km

where M= mass of the moon = 7.3x10^22kg

P=period

G=newtonian gravatational constant= 6.67x10^-11

To find the Period solve for P

P = √[(4π²/G M)xd³]

P=√(4 π²/6.67x10^-22*7.3x10^22kg) x (1.838x10^6m)³]

= 3.6*10^18s

0.863333

86%

hope this

b mirrors submit a iisner? e. for which optical devices does d sometimes have a positive value? o a a and b o b. a and c o c. b and d 0 d. b and cb