How do the Sun's attraction and the Earth's rotation affect the readings of a spring scale that measures the weight of a body at the equator at noon and midnight? Consider that Earth's axis of rotation is perpendicular to the plane of the orbit.
Decision:
A body suspended on a spring scale is subject to the following forces: the Earth's gravity $mg$, the Sun's gravity $G \frac{M_{C}m}{R^{\prime \: 2}}$ ($R^{\prime}$ is the distance from the body to the center of the Sun), and the spring tension P (the reading of the scale). Under the action of these forces, the body experiences accelerations associated with the Earth's rotation around its axis
$a_{1} =\frac{4 \pi^{2}r}{\tau^{2}}=\omega^{2}r$.
and the Earth's rotation around the Sun
$a_{2}=\frac{4 \pi^{2}r}{T^{2}}=G \frac{M_{C}}{R^{2}}$ $T=1 \: y$
Here $R$ is the radius of the Earth's orbit and $r$ is the radius of the Earth. At noon (index 1) and at midnight (index 2) the body, the center of the Earth and the Sun are on the same straight line, so all forces and accelerations are directed along the same axis. By Newton's II law of motion, we get:
$mg-P_{1}- G \frac{M_{C}m}{R^{2}_{1}}=m \left ( \omega^{2}r - m G \frac{M_{C}}{R^{2}} \right )$, (1)
$mg-P_{2}- G \frac{M_{C}m}{R^{2}_{2}}=m \left ( \omega^{2}r + m G \frac{M_{C}}{R^{2}} \right )$, (2)
where $R_{1}=R-r, R_{2}=R+r$, are the distances from the body to the center of the Sun at noon and midnight, respectively. From these equations, the values of $P_{1}$ and $P_{2}$ can be obtained. When performing, we should take into account the smallness of the Earth's radius compared to the radius of its orbit:
$\frac{1}{R \pm r}^{2} \simeq \frac{1}{R^{2}} \left( 1 \mp \frac{2r}{R} \right)$.
The result of the calculation:
$P_{1} \approx P_{2} \approx m \left( g- \frac{4 \pi^{2}r }{\tau^{2}r }{\tau^{2}} - 2\frac{4 \pi^{2}r}{T^{2}} \right)$ (3)
It can be seen that the last summand in the right-hand side of equality (3) can be neglected ($T \gg \tau$).
Answer:
The corrections to the weight of the body due to the rotation of the Earth on its axis around the Sun at noon and at midnight are the same and equal to
$\Delta P \approx -4 \pi^{2} mr \frac{1}{\tau^{2}} \approx - 0.0034 mg$.
Decision:
A body suspended on a spring scale is subject to the following forces: the Earth's gravity $mg$, the Sun's gravity $G \frac{M_{C}m}{R^{\prime \: 2}}$ ($R^{\prime}$ is the distance from the body to the center of the Sun), and the spring tension P (the reading of the scale). Under the action of these forces, the body experiences accelerations associated with the Earth's rotation around its axis
$a_{1} =\frac{4 \pi^{2}r}{\tau^{2}}=\omega^{2}r$.
and the Earth's rotation around the Sun
$a_{2}=\frac{4 \pi^{2}r}{T^{2}}=G \frac{M_{C}}{R^{2}}$ $T=1 \: y$
Here $R$ is the radius of the Earth's orbit and $r$ is the radius of the Earth. At noon (index 1) and at midnight (index 2) the body, the center of the Earth and the Sun are on the same straight line, so all forces and accelerations are directed along the same axis. By Newton's II law of motion, we get:
$mg-P_{1}- G \frac{M_{C}m}{R^{2}_{1}}=m \left ( \omega^{2}r - m G \frac{M_{C}}{R^{2}} \right )$, (1)
$mg-P_{2}- G \frac{M_{C}m}{R^{2}_{2}}=m \left ( \omega^{2}r + m G \frac{M_{C}}{R^{2}} \right )$, (2)
where $R_{1}=R-r, R_{2}=R+r$, are the distances from the body to the center of the Sun at noon and midnight, respectively. From these equations, the values of $P_{1}$ and $P_{2}$ can be obtained. When performing, we should take into account the smallness of the Earth's radius compared to the radius of its orbit:
$\frac{1}{R \pm r}^{2} \simeq \frac{1}{R^{2}} \left( 1 \mp \frac{2r}{R} \right)$.
The result of the calculation:
$P_{1} \approx P_{2} \approx m \left( g- \frac{4 \pi^{2}r }{\tau^{2}r }{\tau^{2}} - 2\frac{4 \pi^{2}r}{T^{2}} \right)$ (3)
It can be seen that the last summand in the right-hand side of equality (3) can be neglected ($T \gg \tau$).
Answer:
The corrections to the weight of the body due to the rotation of the Earth on its axis around the Sun at noon and at midnight are the same and equal to
$\Delta P \approx -4 \pi^{2} mr \frac{1}{\tau^{2}} \approx - 0.0034 mg$.