Answer to Question #112864 in Mechanics | Relativity for Clara

Question

As we discussed, the Earth's orbit is ever so slightly eccentric. At its closest approach the Earth is 1.471x109m from the sun. At its furthest point the Earth is 1.521x109m from the sun. At it's furthest point the speed of the Earth in its orbit is 2.929x104m/s. Calculate its speed at the closest approach. Express your answer in m/s

Answer 1

Explanation

• When an object orbit another stationary one (usually planets) on an elliptical path, the force between them which keep the moving object in a circular path is perpendicular to the path.
• So it does not impose any external torque on the moving object hence its angular momentum is conserved.
• Equation for rotational momentum can be written both using angular velocity & the instantaneous linear velocity.

Calculations

$\qquad \qquad \begin{aligned} \small I_1\omega_1 &= \small I_2\omega_2\\ \small mv_1r_1 &= \small mv_2r_2 \cdots\cdots(\text{substitute I and} \,\omega\,\text{with}\, v = r\omega \,\,\text{and} \,\, I= mr^2)\\ \small \cancel {m}v_1r_1 &= \small \cancel{m}v_2r_2\\ \small \bold {v_{furthest}r_{furthest}} &= \small \bold{v_{closest}r_{closest}} \end{aligned}$

• Now the required data is given & the required velocity could be calculated

$\qquad \qquad \begin{aligned} \small 2.929*10^4ms^{-1}\times1.521*10^9m &= \small v_{closest}\times1.471*10^9m\\ \small \bold {v_{closest}} &= \small \bold{3.029*10^4ms^{-1}} \end{aligned}$