Space Lab Report
By looking at both graphs we can see that from □(→┬AB ) the spaceship is accelerating from 0 to 0.95c until it reaches the Asteroid belt. The asteroid belt is located at 1.7AU far from earth , Then 95% of the speed of light can be written as 2.85X〖10〗^8 m/s and one AU equals 1.496×〖10〗^11 mt, to calculate how much time the spaceship takes to reach B we have
(1.7(1.496x〖10〗^11 m))/(1/2((2.85×〖10〗^8 m)/s))=~1778s or approximately 30 minutes. Whereas from C to D the spaceship desaccelerates at a distance of 1.7AU Before reaching Kepteyn b, taking this into account from □(→┬CD ) it will also take ~ 30min.
Having say that, the distance travel through →┬BC equals to the distance from Earth to Kepteyn b minus the distance it takes for the spaceship to accelerate and deaccelerate. So as Diagram 1 shows the space shuttle will travel …show more content…
(809471.7AU-2(1.7AU)= 809468.3 AU, and according to diagram 2 is going to do it at a constant velocity of 0.95c. Thus the spaceship will take ((809471.7-2(1.7)(1.496×〖10〗^11 m))/(((2.85×〖10〗^8 m)/s))= 424899852 seconds, and in years 424899852/(60×60×24×365.25)=
3.46426381 years.
to travel between B and C.
Thirdly, □(→┬DE ) Represents the 5 years in which the Astronouts will spend collecting data on the planet, that’s why theres no acceleration, velocity or distance travel, only time. After that, the astronouts will repeat the same trip back to Earth, starting with the acceleration through □(→┬EF ) leaving Kepteyn b, to then travel at a constant speed from □(→┬FG ) untill they reach the asteroid belt and desaccelerate through □(→┬GH ) and arriving Earth safe and sound.
Neverthless it is important to mention that Diagram 1 displays a negative displacement because the spaceship is returning to point 0 (Earth). As a consequence the velocity of the spaceship will be also negative (diagram 2) because a negative displacement divided by time equals a negative velocity ((-)Displacement)/((+)Time)=(-)v and therefore v=-0.95c
Nonethless time remains positive, (+)time= ((-)Displacement)/((-)v) the division between two negative numbers equal a positive value. So the displacement and velocity returning Earth will be the same magnitudes but in negative values and the time taken will
thesesame.
(1.7(1.496x〖10〗^11 m))/(1/2((2.85×〖10〗^8 m)/s))=~1778s or approximately 30 minutes. Whereas from C to D the spaceship desaccelerates at a distance of 1.7AU Before reaching Kepteyn b, taking this into account from □(→┬CD ) it will also take ~ 30min.
Having say that, the distance travel through →┬BC equals to the distance from Earth to Kepteyn b minus the distance it takes for the spaceship to accelerate and deaccelerate. So as Diagram 1 shows the space shuttle will travel …show more content…
(809471.7AU-2(1.7AU)= 809468.3 AU, and according to diagram 2 is going to do it at a constant velocity of 0.95c. Thus the spaceship will take ((809471.7-2(1.7)(1.496×〖10〗^11 m))/(((2.85×〖10〗^8 m)/s))= 424899852 seconds, and in years 424899852/(60×60×24×365.25)=
3.46426381 years.
to travel between B and C.
Thirdly, □(→┬DE ) Represents the 5 years in which the Astronouts will spend collecting data on the planet, that’s why theres no acceleration, velocity or distance travel, only time. After that, the astronouts will repeat the same trip back to Earth, starting with the acceleration through □(→┬EF ) leaving Kepteyn b, to then travel at a constant speed from □(→┬FG ) untill they reach the asteroid belt and desaccelerate through □(→┬GH ) and arriving Earth safe and sound.
Neverthless it is important to mention that Diagram 1 displays a negative displacement because the spaceship is returning to point 0 (Earth). As a consequence the velocity of the spaceship will be also negative (diagram 2) because a negative displacement divided by time equals a negative velocity ((-)Displacement)/((+)Time)=(-)v and therefore v=-0.95c
Nonethless time remains positive, (+)time= ((-)Displacement)/((-)v) the division between two negative numbers equal a positive value. So the displacement and velocity returning Earth will be the same magnitudes but in negative values and the time taken will
thesesame.