Respuesta :
Answer:
v = 7121.3 m/s
Explanation:
As we know that the centripetal force for the space shuttle is due to gravitational force of earth due to which it will rotate in circular path with constant speed
so here we will have
[tex]\frac{mv^2}{r} = \frac{GMm}{r^2}[/tex]
here we know that
r = orbital radius = 6370 km + 1482 km
[tex]r = 7.852 \times 10^6 m[/tex]
also we know that
[tex]M = 5.97 \times 10^{24} kg[/tex]
now we will have
[tex]v^2 = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{7.852 \times 10^6}[/tex]
[tex]v^2 = 5.07 \times 10^7[/tex]
[tex]v = 7121.3 m/s[/tex]
The speed of the space shuttle that orbited the earth at an altitude of 1482km will be [tex]V=7121.3\dfrac{m}{s}[/tex]
What will be the speed of a space shuttle that orbited Earth at an altitude of 1482 km?
As we know that the centripetal force for the space shuttle is due to the gravitational force of the earth due to which it will rotate in a circular path with constant speed
so here we will have
[tex]\dfrac{mv^2}{r} = \dfrac{GMm}{r^2}[/tex]
[tex]V^2=\dfrac{GM}{r}[/tex]
here we know that
[tex]\rm r= orbital \ radius =6370+1482=7852 \ km[/tex]
[tex]r=7.852\times10^6\ m[/tex]
mass of earth [tex]M=5.97\times10^{24}\ kg[/tex]
Gravitational constant [tex]G=6.67\times10^{-11}[/tex]
By putting all the values we get
[tex]V^2=\dfrac{(6.67\times10^{-11} )(5.97\times10^{24})}{7.852\times10^{6}}[/tex]
[tex]V^2=5.07\times10^7[/tex]
[tex]V=7121.3 \dfrac{m}{s}[/tex]
Thus the speed of the space shuttle that orbited the earth at an altitude of 1482km will be [tex]V=7121.3\dfrac{m}{s}[/tex]
