Spacecraft Essay

The asteroid ceres has an orbital speed around the sun of 17.8926km/s

Ceres orbital speed(S) is 17.8926 km/s

The distance from the earth to the asteroid ceres at time (T) = 0 is 3.00 x 108

Speed of spacecraft is 8.13876689 km/s

(a)

Time is given by Distance travelled divide by the speed

Time = Distance travelled/ Speed

T = D/S

T = {3.00  108}/ 8.13876689

T = 0.368606208  108

Seconds

The spacecraft will take 0.368606208  108 Seconds to get to the Ceres.

(b)            

Distance covered by the Ceres during the time is given speed multiplied by time

In this case, time distance is expressed as:

Distance = Speed  Time

D = ST

D = {17.8926km/s }{0.368606208  108 }

                                                D = 6.5953234355  108 Km

Ceres moved a distance of 65.953234373  108

Km during the time

(c)

When the spacecraft speed is 8.13 Km/s

Time = D/S

Time = {3.00  108}/ 8.13

Time = 0.36900369  108

Time = 0.36900369  108

Seconds

When the spacecraft speed is 8.13 Km/s, time taken by the spacecraft to reach ceres becomes 0.36900369  108

Seconds

(d)

Distance is given by Speed (S) multiplied by Time (T)

Distance = Speed  Time

D = ST

= 17.8926  0.36900369  108

6.6024354237  108

Km

(e)   [{6.6024354237  108

}- {6.5953234355  108 }] Km

=7.1119882 105

Km

Difference in the distance is 7.1119882 105 Km

By rounding up the velocities, the spacecraft will miss ceres by 7.1119882 105

Km

(f)

Distance from the earth to the moon is 384,400 km

Therefore {711198.82/ 384400} = 1.85

Almost 2 times.