Use Kepler's third law to determine how many days it takes a spacecraft to travel in an elliptical orbit from a point 6 590 km from the Earth's center to the Moon, 385 000 km from the Earth's center.

1.363×10^15 seconds

Explanation:

The spaceship travels an elliptical orbit from a point of 6590km from the earth center to the moon and 38500km from the earth center.

To calculate the time taken from Kepler's third Law :

T^2 = ( 4π^2/GMe ) r^3

Where Me is the mass of the earth

r is the average distance travel

G is the universal gravitational constant. = 6.67×10-11 m3 kg-1 s-2

π = 3.14

Me = mass of earth = 5.972×10^24kg

r =( r minimum + r maximum)/2 1

rmin = 6590km

rmax = 385000km

From equation 1

r = (6590+385000)/2

r = 391590/2

r = 195795km

From T^2 = ( 4π^2/GMe ) r^3

T^2 = (4 × 3.14^2/ 6.67×10-11 × 5.972×10^24) × 195795^3

= ( 4×9.8596/ 3.983×10^14 ) × 7.5059×10^15

= 39.4384/ 3.983×10^14 ) × 7.5059×10^15

= (9.901×10^14) × 7.5059×10^15

T^2 = 7.4321× 10^30

T =√7.4321× 10^30

T = 2.726×10^15 seconds

The time for one way trip from Earth to the moon is :

∆T = T/2

= 2.726×10^15 /2

= 1.363×10^15 secs

correct it is c.) b b

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