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Chapter 12: Problem 17

Is the orbital speed of the Earth when it is closest to the Sun greater than, less than, or equal to the orbital speed when it is farthest from the Sun? Explain.

Short Answer

Expert verified
Answer: The orbital speed of Earth when it is closest to the Sun (perihelion) is greater than when it is farthest from the Sun (aphelion).

Step by step solution

01

Understand the concept of orbital speed

Orbital speed is the speed at which an object (in this case, the Earth) moves in its orbit around another object (the Sun). The primary factor affecting the orbital speed of an object is the gravitational force acting between the objects, which depends on their masses and the distance between their centers.
02

Apply Kepler's second law

Kepler's second law, also known as the Law of Equal Areas, states that a line joining a planet and its star sweeps out equal areas in equal periods of time. This law implies that when the distance between the planet and the star is reduced, the planet moves faster to cover the same area in the same time.
03

Determine the orbital speed at perihelion and aphelion

We need to find the Earth's orbital speed when it is closest to the Sun (perihelion) and when it is farthest from the Sun (aphelion). According to Kepler's second law, Earth moves faster at perihelion than at aphelion because of the different distances between Earth and Sun.
04

Compare the orbital speeds at perihelion and aphelion

Since Earth moves faster when it is closer to the Sun (perihelion), we can conclude that the orbital speed of Earth when it is closest to the Sun is greater than when it is farthest from the Sun.

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Most popular questions from this chapter

The more powerful the gravitational force of a planet, the greater its escape speed, \(v,\) and the greater the gravitational acceleration, \(g\), at its surface. However, in Table 12.1 , the value for \(v\) is much greater for Uranus than for Earth - but \(g\) is smaller on Uranus than on Earth! How can this be?

a) What is the total force on \(m_{1}\) due to \(m_{2}, m_{3},\) and \(m_{4}\) if all four masses are located at the corners of a square of side \(a\) ? Let \(m_{1}=m_{2}=m_{3}=m_{4}\). b) Sketch all the forces acting on \(m_{1}\).

Standing on the surface of a small spherical moon whose radius is \(6.30 \cdot 10^{4} \mathrm{~m}\) and whose mass is \(8.00 \cdot 10^{18} \mathrm{~kg}\) an astronaut throws a rock of mass 2.00 kg straight upward with an initial speed \(40.0 \mathrm{~m} / \mathrm{s}\). (This moon is too small to have an atmosphere.) What maximum height above the surface of the moon will the rock reach?

A satellite is in a circular orbit around a planet. The ratio of the satellite's kinetic energy to its gravitational potential energy, \(K / U_{\mathrm{g}}\), is a constant whose value is independent of the masses of the satellite and planet, and of the radius and velocity of the orbit. Find the value of this constant. (Potential energy is taken to be zero at infinite separation.)

Express algebraically the ratio of the gravitational force on the Moon due to the Earth to the gravitational force on the Moon due to the Sun. Why, since the ratio is so small, doesn't the Sun pull the Moon away from the Earth?
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