Analyzing the James Webb Space Telescope's Orbit
A detailed analysis of the James Webb Space Telescope's orbit using gravitational physics and approximations.
Problem Statement
The James Webb Space Telescope (JWST) orbits in a unique configuration, maintaining its position along the Sun-Earth line, farther from the Sun than the Earth. In a simplified model:
- Assume the Earth is in a circular orbit around the Sun with a period $T = 365 \, \text{days}$.
- The Sun has a mass $M$ much greater than the mass of the Earth $m$, which is itself much greater than the mass of the telescope $\mu$.
- The distance between the Sun and the Earth is $L$, and the distance between the Earth and the telescope is $x$.
- It is given that $x \ll L$.
Tasks
- (a) Derive a formula for the mass of the Earth $m$, expressed in terms of the gravitational constant $G$, the gravitational acceleration at the surface of the Earth $g$, and the radius of the Earth $R$.
- (b) Derive a formula for $x$, expressed in terms of $G$, $m$, and $T$, using appropriate approximations for small values of $x/L$.
- (c) Transform the formula for $x$ to express it in terms of $T$, $g$, and $R$.
Solution
(a) Derive the Mass of the Earth
At the Earth’s surface, the gravitational field strength $g$ is related to the Earth’s mass $m$ by:
\[g = \frac{G m}{R^2}.\]Rearranging for $m$:
\[m = \frac{g R^2}{G}.\]This provides a relationship between $g$, $R$, $G$, and $m$.
(b) Derive a Formula for $x$ in Terms of $G$, $m$, and $T$
Gravitational Forces and Accelerations
We analyze the forces acting on the telescope and the Earth, assuming circular orbits.
1. Earth’s Orbit (Including Telescope’s Gravitational Force)
The Earth experiences two gravitational forces:
- The force from the Sun, $\frac{G M m}{L^2}$,
- The force from the telescope, $\frac{G m \mu}{x^2}$.
The net radial force acting on the Earth is:
\[\frac{G M m}{L^2} - \frac{G m \mu}{x^2} = m \frac{v_m^2}{L},\]where $v_m$ is the Earth’s orbital speed. Substituting $v_m = \frac{2 \pi L}{T}$:
\[\frac{G M}{L^2} - \frac{G \mu}{x^2} = \frac{4 \pi^2 L}{T^2}.\]Justification for Neglecting the Telescope’s Contribution
The term $\frac{G \mu}{x^2}$ is small compared to $\frac{G M}{L^2}$ because $\mu \ll M$ and $x \ll L$. The ratio of the telescope’s contribution to the Sun’s force is:
\[\frac{\frac{G \mu}{x^2}}{\frac{G M}{L^2}} = \frac{\mu L^2}{M x^2}.\]Since $\mu \ll M$ and $x \ll L$, this ratio is negligible. Mathematically, we retain $\frac{G \mu}{x^2}$ in the equation for rigor, but as $x \to 0$ and $\mu \ll M$, this term vanishes. Hence, we approximate:
\[\frac{G M}{L^2} \approx \frac{4 \pi^2 L}{T^2}.\]This implies:
\[\frac{G M}{L^3} = \frac{4 \pi^2}{T^2}.\]Telescope’s Orbit: The telescope experiences gravitational forces from both the Sun and the Earth. The net radial force is:
\[\frac{G M \mu}{(L+x)^2} + \frac{G m \mu}{x^2} = \mu \frac{v_\mu^2}{L+x},\]where $v_\mu$ is the orbital speed of the telescope. Substituting $v_\mu = \frac{2 \pi (L+x)}{T}$:
\[\frac{G M}{(L+x)^2} + \frac{G m}{x^2} = \frac{4 \pi^2 (L+x)}{T^2}.\]
Approximations for Small $x/L$
Given $x \ll L$, approximate terms as follows:
- $\frac{1}{(L+x)^2} \approx \frac{1}{L^2} \left(1 - \frac{2x}{L}\right) = \frac{1}{L^2} - \frac{2x}{L^3}$.
Substituting into $\frac{G M}{(L+x)^2}$:
\[\frac{G M}{(L+x)^2} \approx \frac{G M}{L^2} - \frac{2 G M x}{L^3}.\]Expanding $\frac{4 \pi^2 (L+x)}{T^2}$ exactly:
\[\frac{4 \pi^2 (L+x)}{T^2} = \frac{4 \pi^2 L}{T^2} + \frac{4 \pi^2 x}{T^2}.\]Using these results:
\[\frac{G M}{L^2} - \frac{2 G M x}{L^3} + \frac{G m}{x^2} = \frac{4 \pi^2 L}{T^2} + \frac{4 \pi^2 x}{T^2}.\]Equating the leading terms:
\[\frac{G M}{L^2} = \frac{4 \pi^2 L}{T^2}.\]Substituting this result:
\[- \frac{2 G M x}{L^3} + \frac{G m}{x^2} = \frac{4 \pi^2 x}{T^2}.\]Simplifying further:
\[\frac{G m}{x^2} = \frac{4 \pi^2 x}{T^2} + \frac{2 G M x}{L^3}.\]Since $\frac{G M}{L^3} = \frac{4 \pi^2}{T^2}$:
\[\frac{G m}{x^2} = \frac{4 \pi^2 x}{T^2} + \frac{8 \pi^2 x}{T^2}.\]Combining terms:
\[\frac{G m}{x^2} = \frac{12 \pi^2 x}{T^2}.\]Rearranging for $x$:
\[x^3 = \frac{G m T^2}{12 \pi^2}.\]Thus:
\[x = \sqrt[3]{\frac{G m T^2}{12 \pi^2}}.\](c) Transform $x$ in Terms of $T$, $g$, and $R$
Substitute $m = \frac{g R^2}{G}$ into the formula for $x$:
\[x^3 = \frac{G \left(\frac{g R^2}{G}\right) T^2}{12 \pi^2}.\]Simplify:
\[x^3 = \frac{g R^2 T^2}{12 \pi^2}.\]Thus:
\[x = \sqrt[3]{\frac{g R^2 T^2}{12 \pi^2}}.\]Final Results
Mass of the Earth: \(m = \frac{g R^2}{G}.\)
Distance $x$ in Terms of $G$, $m$, and $T$: \(x = \sqrt[3]{\frac{G m T^2}{12 \pi^2}}.\)
Distance $x$ in Terms of $T$, $g$, and $R$: \(x = \sqrt[3]{\frac{g R^2 T^2}{12 \pi^2}}.\)
Conclusion
This analysis rigorously justifies why the telescope’s gravitational influence is negligible when considering the Earth’s orbit. It also demonstrates how approximations and gravitational physics can provide insight into the James Webb Space Telescope’s stable position in the Sun-Earth system.