Question

The James Webb Space Telescope is positioned around 1.5 million kilometres from the Earth on the side facing away from the Sun. The telescope remains at this distance and orbits around the Sun with the Earth’s orbital velocity.
--Determine the angular velocity ω of the telescope as it orbits around the Sun.

--The centrifugal Fω and gravitational force FG are acting on objects orbiting the Sun: F =Fω−FG. Based on this, how much should the telescope accelerate towards or away from the Sun?

--Why is the orbit of the telescope stable nonetheless? What other forces need to be considered?

Answer 1

The angular velocity depends on the length of the orbit and the orbital

speed of the telescope.

Response:

First question:

• The angular velocity of the telescope is approximately 0.199 rad/s

Second question:

• The telescope should accelerates away by approximately F = 0.0005·m

Third question:

• The pulling force between the Earth and the satellite

What equations can be used to calculate the velocity and forces acting on the telescope?

The distance of the James Webb telescope from the Sun = 1.5 million kilometers from Earth on the side facing away from the Sun

The orbital velocity of the telescope = The Earth's orbital velocity

First question:

$Angular \ velocity = \mathbf{\dfrac{Angle \ turned}{Time \ taken}}$

The orbital velocity of the Earth = 29.8 km/s

The distance between the Earth and the Sun = 148.27 million km

The radius of the orbit of the telescope = 148.27 + 1.5 = 149.77

Radius of the orbit, r = 149.77 million kilometer from the Sun

The length of the orbit of the James Webb telescope = 2 × π × r

Which gives;

r = 2 × π × 149.77 million kilometers ≈ 941.03 million kilometers

Therefore;

$Angular \ velocity = \dfrac{29.8}{941.03}\times 2 \times \pi \approx 0.199$

• The angular velocity of the telescope, ω ≈ 0.199 rad/s

Second question:

Centrifugal force force, $F_{\omega}$ = m·ω²·r

Which gives;

$F_{\omega} = m \cdot \dfrac{28,500^2 \, m^2/s^2}{149.77 \times 10^9 \, m} \approx 0.0054233 \cdot m$

$Gravitational \ force, F_G = \mathbf{G \cdot \dfrac{m_{1} \cdot m_{2}}{r^{2}}}$

Universal gravitational constant, G = 6.67408 × 10⁻¹¹ m³·kg⁻¹·s⁻²

Mass of the Sun = 1.989 × 10³⁰ kg

Which gives;

$F_G = 6.67408 \times 10^{-11} \times \dfrac{1.989 \times 10^{30} \times m}{149.77 \times 10^9} \approx 0.00592 \cdot m$

Which gives;

$F_{\omega}$ < $F_G$, therefore, the James Webb telescope has to accelerate away from the Sun

F = $\mathbf{F_{\omega}}$ - $\mathbf{F_G}$

The amount by which the telescope accelerates away is approximately 0.00592·m - 0.0054233·m ≈ 0.0005·m (away from the Sun)

Third part:

Other forces include;

• The force of attraction between the Earth and the telescope which can contribute to the the telescope having a stable orbit at the given speed.

Learn more about orbital motion here:

brainly.com/question/11069817