Tues. 4/21 and Wed. 4/22 - Universal Gravitation and Orbital Motion Review Part 2

Summary

In this video, the speaker walks through solving an orbital motion problem related to the Hubble Space Telescope. They begin by explaining the relationship between forces, focusing on the gravitational force and centripetal force. The speaker uses these principles to derive the orbital speed equation, taking into account the mass of the Earth and the distance to the telescope. They also highlight the approximation needed due to the disparity in the Earth’s radius and the telescope's height. Finally, the speaker calculates the orbital speed of the Hubble Space Telescope, arriving at a value of 7,580 m/s.

Takeaways

• 😀 The instructor demonstrates solving the orbital speed of the Hubble Space Telescope.
• 😀 The problem involves standard orbital motion variables: V (velocity), R (distance), M (Earth's mass), and m (Hubble's mass).
• 😀 Centripetal force (mv²/r) is set equal to gravitational force (G*M*m/r²) to find orbital speed.
• 😀 The mass of the Hubble Space Telescope cancels out during the calculation.
• 😀 The distance between Earth and the telescope is approximated as the Earth's radius plus the telescope's height.
• 😀 The instructor approximates Earth's radius as 6.4 million meters and the telescope's height as 600,000 meters.
• 😀 The orbital speed formula simplifies to V = √(G*M/R).
• 😀 The gravitational constant G is 6.7 × 10⁻¹¹ N·m²/kg² and Earth's mass is about 6 × 10²⁴ kg.
• 😀 The calculated orbital speed of the Hubble Space Telescope is approximately 7,580 m/s.
• 😀 The instructor notes that solving these problems can be error-prone, acknowledging a personal 20% chance of mistakes.

Q & A

What is the main goal of the problem the speaker is solving?

The speaker is solving for the orbital speed (V) of the Hubble Space Telescope in orbit around the Earth.

Why does the speaker mention the masses of the Earth and the Hubble Space Telescope?

The masses of both the Earth (M) and the Hubble Space Telescope (m) are used in the gravitational force formula to find the orbital speed of the telescope.

What is the significance of the equation mv² / r = G * M * m / r²?

This equation sets the centripetal force (mv² / r) equal to the gravitational force (G * M * m / r²) acting on the Hubble Space Telescope, which allows the speaker to solve for the orbital speed.

Why can the mass of the Hubble Space Telescope (m) be canceled out of the equation?

The mass of the Hubble Space Telescope (m) appears on both sides of the equation and cancels out, simplifying the formula.

What does the speaker mean by 'most of our lowercase r is just the radius of the Earth'?

The speaker explains that the distance between the telescope and the Earth's center is primarily made up of the Earth's radius, since the height of the Hubble Space Telescope above the Earth's surface is relatively small.

How does the speaker estimate the total distance (r) between the telescope and the Earth?

The speaker estimates the total distance (r) as roughly 7 million meters by adding the Earth's radius (about 6.4 million meters) and the altitude of the Hubble Space Telescope (about 600,000 meters).

What are the values used to calculate the orbital speed?

The speaker uses the gravitational constant (G = 6.7 * 10⁻¹¹), the mass of the Earth (M = 6 * 10²⁴), and the total distance (r = 7 million meters) to calculate the orbital speed.

Why does the speaker mention the chance of getting the wrong answer?

The speaker humorously acknowledges the difficulty of solving this type of problem and the possibility of making a mistake, which could lead to an incorrect result.

What is the final calculated orbital speed of the Hubble Space Telescope?

The final orbital speed (V) of the Hubble Space Telescope is approximately 7,580 meters per second (m/s).

What is the speaker's approach to solving the problem, and how does it reflect the complexity of orbital mechanics?

The speaker walks through the process methodically, showing how to apply principles of orbital mechanics (like centripetal force and gravitational force) to find the orbital speed. This reflects the complexity of such problems, which require careful consideration of variables and simplifications.