a) The force of gravity acting on a 490 kilogram spacecraft is 2767.197 newtons.
b) The spacecraft is travelling around the Earth at a speed of approximately 6885.846 meters per second.
a) The force of gravity experimented by the spacecraft is described by the Newton's Law of Gravitation:
[tex]F = \frac{G\cdot m\cdot M}{r^{2}}[/tex] (1)
Where:
- [tex]F[/tex] - Force of gravity, in newtons.
- [tex]G[/tex] - Gravitational constant, in cubic meters per kilogram-square second.
- [tex]m[/tex] - Mass of the spacecraft, in kilograms.
- [tex]M[/tex] - Mass of the Earth, in kilograms.
- [tex]r[/tex] - Distance of the spacecraft with respect to the center of the planet, in meters.
If we know that [tex]G = 6.67\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex], [tex]m = 490\,kg[/tex], [tex]M = 5.972\times 10^{24}\,kg[/tex] and [tex]r = 8.401\times 10^{6}\,m[/tex], then the magnitude of the force of gravity is:
[tex]F = \frac{\left(6.674\times 10^{-11}\,\frac{m^{2}}{kg\cdot s^{2}} \right)\cdot (490\,kg)\cdot (5.972\times 10^{24}\,kg)}{(8.401\times 10^{6}\,m)^{2}}[/tex]
[tex]F = 2767.197\,N[/tex]
The force of gravity acting on a 490 kilogram spacecraft is 2767.197 newtons.
b) The spacecraft experiments a centripetal acceleration, which means that it rotates around the planet at constant speed:
[tex]\frac{G\cdot M}{r^{2}} = \frac{v^{2}}{r}[/tex]
[tex]v = \sqrt{\frac{G\cdot M}{r} }[/tex] (2)
If we know that [tex]G = 6.67\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex], [tex]M = 5.972\times 10^{24}\,kg[/tex] and [tex]r = 8.401\times 10^{6}\,m[/tex], then the magnitude of the force of gravity is:
[tex]v = \sqrt{\frac{\left(6.67\times 10^{-11}\,\frac{m^{2}}{kg\cdot s^{2}\right)\cdot (5.972\times 10^{24}\,kg)} }{8.401\times 10^{6}\,m} }[/tex]
[tex]v \approx 6885.846\,\frac{m}{s}[/tex]
The spacecraft is travelling around the Earth at a speed of approximately 6885.846 meters per second.
We kindly invite to check this question on centripetal acceleration: https://brainly.com/question/1052267
