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Answer: The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.
1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.
2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.
Explanation: To find the answer, we need to know about the different equations of planetary motion.
How to find the initial speed of the rock as it left the astronaut's hand?
- We have the expression for the initial velocity as,
[tex]v=\sqrt{2gh}[/tex]
- Thus, to find v, we have to find the acceleration due to gravity of glob. For this, we have,
[tex]g_g=\frac{GM}{r^2} =\frac{6.67*10^{-11}*7.88*10^{18}}{(6.32*10^4)^2}= 0.132[/tex]
- Now, the velocity will become,
[tex]v=\sqrt{2*0.132*1.44*10^3} =19.46 m/s[/tex]
How to find the speed of the satellite?
- As we know that, by equating both centripetal force and the gravitational force, we get the equation of speed of a satellite as,
[tex]v=\sqrt{\frac{GM}{r} } =\sqrt{\frac{6.67*10^{-11}*7.88*10^{18}}{1.45*10^5} } =3.624km/s[/tex]
Thus, we can conclude that,
1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.
2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.
Learn more about the equations of planetary motion here:
brainly.com/question/28108487
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The tiny planet known as "Glob" has a radius of 6.32× 10^4 meters and a mass of 7.88× 10^18 kg. On Glob's surface, an astronaut launches a rock straight upward. Before falling back down, the rock rises to a maximum height of 1.44×10^3 m above the planet's surface.
1) The rock was moving at 19.46 m/s when it first left the astronaut's palm.
2) A 36.0 kg spacecraft is orbiting the planet Glob in a sphere with a radius of 1.45 105 meters. The satellite is moving at 3.624 km/s at that point.
Understanding the planetary motion equations is necessary in order to determine the solution.
How to determine the rock's original speed when it left the astronaut's hand?
- The starting velocity's expression is as follows:
[tex]V=\sqrt{2gh}[/tex]
- So, in order to determine v, we must determine the acceleration of glob caused by gravity. We already have,
[tex]a=\frac{GM}{r^2} =\frac{6.67*10^{-11}*7.88*10^{18}}{(6.32*10^4)^2} \\a=0.132m/s^2[/tex]
- The velocity will now change to,
[tex]V=\sqrt{2*0.132*1.44*10^3} =19.46m/s[/tex]
How can I determine the satellite's speed?
- As we are aware, the centripetal force and gravitational force are equivalent, and thus leads to the following satellite speed equation:
[tex]v=\sqrt{\frac{GM}{r} } =3,624km/s\\where,\\M=7.88*10^{18}kg[/tex]
Consequently, we can say that
1) The rock was moving at 19.46 m/s when it first left the astronaut's palm.
2) A 36.0 kg spacecraft is orbiting the planet Glob in a sphere with a radius of 1.45 105 meters. The satellite is moving at 3.624 km/s at that point.
Learn more about the planetary motion here:
https://brainly.com/question/28108487
#SPJ4
