(a) What is the linear velocity of the Earth in its orbit…

(a) What is the linear velocity of the Earth in its orbit around the Sun, if the Sun's mass is $m_S = 2 \cdot 10^{30}\ \mathrm{kg}$ and the Earth–Sun distance is $d = 1.5 \cdot 10^{11}\ \mathrm{m}$? (b) What acceleration does the Sun's gravity give at the Earth?

Step 1: Idea

The Earth's orbit is very nearly a circle. For a body to move on a circle something must pull it toward the centre with the centripetal force $mv^2/d$. For the Earth this pull is the Sun's gravity. Setting the two equal gives the speed.

Step 2: (a) Orbital speed

\[ m\frac{v^2}{d} = \gamma\frac{m\,m_S}{d^2}. \]

The Earth's mass $m$ appears on both sides and cancels: the orbital speed at a given distance is the same for any planet, large or small.

\[ v = \sqrt{\frac{\gamma m_S}{d}}, \]

\[ v = \sqrt{\frac{6.67 \cdot 10^{-11} \cdot 2 \cdot 10^{30}}{1.5 \cdot 10^{11}}} = \sqrt{8.89 \cdot 10^8}. \]

\[v \approx 29.8\ \mathrm{km/s}\]

Step 3: (b) The Sun's gravitational acceleration

The Sun's field strength at the Earth's distance equals the acceleration it gives to any body there: \[g_S = \gamma\frac{m_S}{d^2} = \frac{6.67 \cdot 10^{-11} \cdot 2 \cdot 10^{30}}{(1.5 \cdot 10^{11})^2},\]

\[ g_S = \frac{1.334 \cdot 10^{20}}{2.25 \cdot 10^{22}}. \]

\[g_S \approx 5.93 \cdot 10^{-3}\ \mathrm{m/s^2}\]

This is about 1650 times weaker than the Earth's own gravity at its surface. We do not feel it at all, because the Sun gives the same acceleration to us and to the whole Earth: we fall around the Sun together.

Check

From the year: $ 2\pi d/T $ $= 2\pi \cdot 1.5 \cdot 10^{11}/3.156 \cdot 10^7 $ $= 29.9\ \mathrm{km/s}\ \checkmark $

$ v^2/d $ $= (2.98 \cdot 10^4)^2/1.5 \cdot 10^{11} $ $= 5.93 \cdot 10^{-3}\ \mathrm{m/s^2} $ $= g_S\ \checkmark $