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The radius of the Earth's orbit around the Sun (assumed to be circular) is 1.50 X 10^8 km, and...

Question:

The radius of the Earth's orbit around the Sun (assumed to be circular) is 1.50 X 10{eq}^8 {/eq} km, and the Earth travels around this orbit in 365 days.

(a) What is the magnitude of the orbital velocity of the Earth, in m/s?

(b) What is the radial acceleration of the Earth toward the Sun, in m/s^2?

(c) Repeat parts (a) and (b) for the motion of the planet Mercury (orbit radius = 5.79 X 10{eq}^7 {/eq} km, orbital period = 88.0 days)

Orbital Speed

Orbital speed is defined as the speed with which an object orbit around another object like a planet around a star, orbital speed can be calculated as the circumference of the orbital path divided by the orbital period.

{eq}\begin{align} v = \frac{2 \pi r }{T} \end{align} {/eq}

Where r is the radius of the circular orbit and T is the time period.

Answer and Explanation:

Data Given

  • Radius of the Earth's orbit {eq}R_E = 1.50 \times 10^8 \ \rm km = 1.50 \times 10^{11} \ \rm m {/eq}
  • Time take by the earth two complete one circle {eq}T_E = 365 \ \rm days {/eq}

Part A) Orbital velocity of the earth

{eq}\begin{align} v_E = \frac{2 \pi R_E }{T_E} \end{align} {/eq}

{eq}\begin{align} v_E = \frac{2 \pi \times 1.50 \times 10^{11} \ \rm m }{365 \times 24 \times 60 \times 60 \ \rm s} \end{align} {/eq}

{eq}\begin{align} \color{blue}{\boxed{v_E = 2.99 \times 10^4 \ \rm m/s}} \end{align} {/eq}


Part B) The radial acceleration is given by

{eq}\begin{align} a_E = \frac{v_E^2 }{R_E} \end{align} {/eq}

{eq}\begin{align} a_E = \frac{(2.99 \times 10^4 \ \rm m/s)^2 }{1.50 \times 10^{11} \ \rm m} \end{align} {/eq}

{eq}\begin{align} \color{blue}{\boxed{a_E = 5.95 \times 10^{-3} \ \rm m/s^2}} \end{align} {/eq}


  • Radius of the Mercury's orbit {eq}R_M = 5.79 \times 10^7 \ \rm km = 5.79 \times 10^{10} \ \rm m {/eq}
  • Time take by the mercury to complete one circle {eq}T_M = 88 \ \rm days {/eq}

Part C) Orbital velocity of the mercury

{eq}\begin{align} v_M = \frac{2 \pi R_M }{T_M} \end{align} {/eq}

{eq}\begin{align} v_M = \frac{2 \pi \times 5.79 \times 10^{10} \ \rm m }{88 \times 24 \times 60 \times 60 \ \rm s} \end{align} {/eq}

{eq}\begin{align} \color{blue}{\boxed{v_M = 4.785 \times 10^4 \ \rm m/s}} \end{align} {/eq}


The radial acceleration is given by

{eq}\begin{align} a_M = \frac{v_M^2 }{R_M} \end{align} {/eq}

{eq}\begin{align} a_M = \frac{(4.785 \times 10^4 \ \rm m/s)^2 }{5.79 \times 10^{10} \ \rm m} \end{align} {/eq}

{eq}\begin{align} \color{blue}{\boxed{a_M= 3.95 \times 10^{-2} \ \rm m/s^2}} \end{align} {/eq}


Learn more about this topic:

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Kepler's Three Laws of Planetary Motion

from Basics of Astronomy

Chapter 22 / Lesson 12
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