Q.There have been suggestions that the value of the gravitational constant $G$ becomes smaller when considered over very large time period (in billions of years) in the future. If that happens, for our earth, (Note: more than one of the given options may be correct.)

Variation of Gravity

If the gravitational constant $G$ decreases over billions of years, the gravitational force binding Earth to the Sun weakens. The key is to recognize that orbital mechanics depend on a balance between gravitational attraction and the orbital motion established under the original value of $G$.

Step 1: Earth's orbital speed was set when $G$ had its current value. The centripetal force required for a circular orbit is $\frac{mv^2}{r} = \frac{GMm}{r^2}$, which gives $v = \sqrt{\frac{GM}{r}}$.

Step 2: If $G$ decreases while Earth retains its tangential velocity (by inertia), the gravitational pull becomes insufficient to maintain the circular orbit. The planet now moves too fast for the weaker gravitational field.

Step 3: With inadequate centripetal force, Earth spirals outward in an open, non-closed trajectory. Over sufficient time, the orbit becomes unbound and Earth escapes the solar system. …

Why a changing $G$ matters

The gravitational constant $G$ appears in every gravitational interaction. For a planet orbiting the Sun, two fundamental quantities are conserved as long as the force law remains unchanged: angular momentum $L$ and energy $E$. But if $G$ varies slowly with time, the system is no longer truly conservative in the usual sense—energy and orbital parameters must adjust.

The key insight is that angular momentum $L = m v r$ is more robust than energy when $G$ changes adiabatically (very slowly). The orbital angular momentum depends on the instantaneous state of motion and is conserved even as $G$ drifts, because no external torque acts on the Earth–Sun system. Energy, however, is tied directly to $G$ through the potential $U = -\frac{G M m}{r}$, so as $G$ decreases, the binding weakens.

For a circular orbit, the balance between centripetal force and gravity gives

The angular momentum is $L = m v r = m \sqrt{G M r}$. If $L$ is conserved while $G$ decreases, the orbital radius $r$ must increase to keep $L$ constant:

As $G \to G - \delta G$ (with $\delta G > 0$), we have $r \to r + \delta r$ with $\delta r > 0$. The orbit spirals outward.

Step-by-step analysis of each option

1. Option (A): Nothing will change.

   This is clearly false. A decrease in $G$ directly weakens the Sun's gravitational pull. The orbital radius must increase to conserve angular momentum, so the Earth's orbit cannot remain unchanged.

2. Option (B): We will become hotter after billions of years.

   Temperature on Earth is governed primarily by solar radiation received, which scales as $1/r^2$ (inverse-square law). As Earth moves to a larger orbital radius $r$, the solar flux decreases, so Earth would receive less energy per unit area and tend to become cooler, not hotter. This option is incorrect.

3. Option (C): We will be going around but not strictly in closed orbits.

   In a static potential (constant $G$), Kepler orbits are closed ellipses. But if $G$ decreases continuously, the potential itself is time-dependent. The orbit at any instant resembles an ellipse, but the semi-major axis and eccentricity evolve secularly—the trajectory is a spiral, not a closed curve. After one "orbit," Earth does not return to the same point in phase space. This is correct.

4. Option (D): After sufficiently long time we will leave the solar system.

   The total mechanical energy of the orbit is

   $E = \frac{1}{2} m v^2 - \frac{G M m}{r} = -\frac{G M m}{2r} \quad \text{(for a circular orbit)}.$ …

Step 1: Earth's orbital speed set by current G: v=sqrt(GM_Sun/r). Step 2: as G slowly decreases while Earth's inertia persists, gravity becomes too weak for the existing orbit. Step 3: orbit radius drifts outward over time -- (c). Step 4: eventually orbit becomes unbound -- Earth could leave the solar system -- (d). S …