Question

The small spherical planet called "Glob" has a mass of 7.88×10^18 kg and a radius of 6.32×10^4 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×10^3 m, above the surface of the planet, before it falls back down.
1. What was the initial speed of the rock as it left the astronaut's hand? (Glob has no atmosphere, so no energy is lost to air friction. G = 6.67×10^-11 Nm2/kg2.)

2. A 36.0 kg satellite is in a circular orbit with a radius of 1.45×10^5 m around the planet Glob. Calculate the speed of the satellite.

Answer 1

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:

                                $V=\sqrt{2gh}$

• So, in order to determine v, we must determine the acceleration of glob caused by gravity. We already have,

                     $a=\frac{GM}{r^2} =\frac{6.67*10^{-11}*7.88*10^{18}}{(6.32*10^4)^2} \\a=0.132m/s^2$

• The velocity will now change to,

                   $V=\sqrt{2*0.132*1.44*10^3} =19.46m/s$

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:

                         $v=\sqrt{\frac{GM}{r} } =3,624km/s\\where,\\M=7.88*10^{18}kg$

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:

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Answer 2

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,

                           $v=\sqrt{2gh}$

• Thus, to find v, we have to find the acceleration due to gravity of glob. For this, we have,

                       $g_g=\frac{GM}{r^2} =\frac{6.67*10^{-11}*7.88*10^{18}}{(6.32*10^4)^2}= 0.132$

• Now, the velocity will become,

                        $v=\sqrt{2*0.132*1.44*10^3} =19.46 m/s$

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,

                       $v=\sqrt{\frac{GM}{r} } =\sqrt{\frac{6.67*10^{-11}*7.88*10^{18}}{1.45*10^5} } =3.624km/s$

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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