Halting Earth's Rotation with a Passing Object

Question: How large would an object need to be, if it passed the earth at 500,000 miles, to stop the earths rotation?

To halt Earth's rotation, an external object would need to exert enough torque to counteract the Earth's angular momentum. The angular momentum of a rotating solid object is given by:

\[ L = I \omega \]

Where:

• \( L \) is the angular momentum
• \( I \) is the moment of inertia of the object
• \( \omega \) is the angular velocity of the object

For a solid sphere like Earth, the moment of inertia is:

\[ I = \frac{2}{5} m r^2 \]

Where:

• \( m \) is the mass of the Earth, approximately \( 5.972 \times 10^{24} \) kg
• \( r \) is the radius of the Earth, approximately \( 6.371 \times 10^6 \) meters

The Earth completes one rotation in about 24 hours, which translates to an angular velocity (\( \omega \)) of:

\[ \omega = \frac{2\pi}{24 \times 60 \times 60} \]

Given these values, we can compute the Earth's angular momentum:

\[ L_{\text{earth}} = I \omega = \frac{2}{5} m r^2 \omega \]

To halt Earth's rotation, an external object passing by would need to exert a torque to counteract this angular momentum. The torque exerted by the external object can be estimated as:

\[ \tau = r_{\text{distance}} \times F \]

Where:

• \( r_{\text{distance}} \) is the distance from the center of the Earth to the line of action of the force (which would be \( 500,000 \) miles plus Earth's radius)
• \( F \) is the gravitational force exerted by the external object on the Earth

Using Newton's law of universal gravitation:

\[ F = \frac{G \times m \times m_{\text{object}}}{{r_{\text{distance}}}^2} \]

Where:

• \( G \) is the gravitational constant, \( 6.674 \times 10^{-11} \) N(m^2)/(kg^2)
• \( m_{\text{object}} \) is the mass of the passing object

Equating the two torques:

\[ r_{\text{distance}} \times \frac{G \times m \times m_{\text{object}}}{{r_{\text{distance}}}^2} = \frac{2}{5} m r^2 \omega \]

From this equation, the mass of the passing object required to halt Earth's rotation is approximately:

\[ m_{\text{object}} \approx 1.435 \times 10^{28} \text{kg} \]

Considerations:

This is a theoretical calculation, and in practice, such a scenario is improbable for various reasons:

1. Scale of the Object: The mass calculated is many times greater than the mass of the Earth itself, implying an object of stellar proportions.
2. Gravitational Consequences: An object of this magnitude passing so close to Earth would have other significant gravitational effects, including altering Earth's orbit or even pulling the Earth out of its current orbit entirely.
3. Tidal Forces: The enormous gravitational forces exerted by such a massive object would generate immense tidal forces on the Earth, potentially leading to catastrophic geological events.
4. Temporal Duration: The calculation also assumes that the passing object exerts its gravitational influence instantaneously to counteract Earth's angular momentum. In reality, the duration of the interaction would play a significant role.

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