Question

calculate the amgnitued of the cetripetal force acting on earth as it orbits the sun, assuming a circular orbit and an orital speed of 3.00 times 10^4 meters per second

Answer 1

Approximately
`3.59* 10^(22)\; {\rm N}` (assuming that the orbit is circular.)

Step-by-step explanation:

When an object is in a circular orbit of radius
`r` with a (linear) speed of
`v`, the centripetal acceleration of this object would be
`a = (v^(2) / r)`. If the mass of the object is
`m`, magnitude of the net force on this object would be
`F = m\, a = (m\, v^(2) / r)`.

Look up the mass of the Earth:
`m \approx 5.9722* 10^(24)\; {\rm kg}`.

Look up the radius of the orbit (mean distance between the Earth and the Sun):
`r \approx 1.4960* 10^(11)\; {\rm m}` (one Astronomical Unit.)

It is given that the linear orbital speed is
`v = 3.00 * 10^(4)\; {\rm m\cdot s^(-1)}`.

Therefore, magnitude of the centripetal force (net force) on the Earth would be:

`\begin{aligned}F &= m\, a = (m\, v^(2))/(r) \\ &\approx \frac{(5.9722 * 10^(24)\; {\rm kg})\, (3.00 * 10^(4)\; {\rm m\cdot s^(-1)})^(2)}{1.4960 * 10^(11)\; {\rm m}} \\ &\approx 3.59 * 10^(22)\; {\rm N} \end{aligned}`.