By considering the conservation of energy, the speed V of the spacecraft when it eventually crash into the earth and its distance from the center of the earth is r are 11172.6 m/s and 3294.6 m/s respectively.
Given that, r is at infinity when a spacecraft has run out of fuel and its kinetic energy is zero. if only the gravitational force of the earth were to act on the spacecraft would eventually crash into the earth.
Where
Mass of the earth [tex]m_{e}[/tex] = 5.97 × [tex]10^{24}[/tex] kg
Radius of the earth [tex]R_{e}[/tex] = 6.38 × [tex]10^{6}[/tex] m
G = 6.67 x [tex]10^{-11}[/tex]N[tex]m^{2}kg^{-2}[/tex]
a.) Since the spacecraft is primarily moving through the near vacuum of space, the speed V of the spacecraft when it eventually crash into the earth can be calculated by considering the conservation of energy.
The total initial energy = total final energy.
[tex]E_{i} = E_{f}[/tex]
[tex]K.E_{i}[/tex] + [tex]U_{i}[/tex] = [tex]K.E_{f}[/tex] + [tex]U_{f}[/tex]
where [tex]K.E_{i}[/tex] = 0
-G[tex]m_{e}[/tex]m/(r + [tex]R_{e}[/tex]) = 1/2m[tex]V^{2}[/tex] + ( -G[tex]m_{e}[/tex]m)/[tex]R_{e}[/tex]
mass of the spacecraft will cancel out
G[tex]m_{e}[/tex]/(r + [tex]R_{e}[/tex]) = 1/2[tex]V^{2}[/tex] + ( -G[tex]m_{e}[/tex])/[tex]R_{e}[/tex]
G[tex]m_{e}[/tex]/(r + [tex]R_{e}[/tex]) = 0 since r is at infinity.
0 = 1/2[tex]V^{2}[/tex] + ( -G[tex]m_{e}[/tex])/[tex]R_{e}[/tex]
Substitute all the parameters into the equation above.
0 = 0.5[tex]V^{2}[/tex] + ( -6.67 x [tex]10^{-11}[/tex] x 5.97 x [tex]10^{24}[/tex] )/6.38 x [tex]10^{6}[/tex]
0.5[tex]V^{2}[/tex] = 62,413,636.36
[tex]V^{2}[/tex] = 62,413,636.36 / 0.5
[tex]V^{2}[/tex] = 124,827,272.7
V = [tex]\sqrt{124,827,272.7}[/tex]
V = 11172.6 m/s
b.) Given that the distance from the earth r = x([tex]R_{e}[/tex])
That is,
r = 11.5 x 6.38 x [tex]10^{6}[/tex]
r = 73,370,000 m
To find the spacecraft's speed when its distance from the center of the earth is r, we will use the same formula. That is,
[tex]K.E_{i}[/tex] + [tex]U_{i}[/tex] = [tex]K.E_{f}[/tex] + [tex]U_{f}[/tex]
0 = 1/2[tex]V^{2}[/tex] + ( -G[tex]m_{e}[/tex])/r
Substitute all the parameters into the formula
0 = 0.5[tex]V^{2}[/tex] + ( -6.67 x [tex]10^{-11}[/tex] x 5.97 x [tex]10^{24}[/tex] )/ 73,370,000
0.5[tex]V^{2}[/tex] = 5427272.727
[tex]V^{2}[/tex] = 10854545.45
V = [tex]\sqrt{10854545.45}[/tex]
V = 3294.6 m/s
Therefore, the speed V of the spacecraft when it eventually crash into the earth and its distance from the center of the earth is r are 11172.6 m/s and
3294.6 m/s respectively.
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