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For every force there is an equal force in the opposite direction on another body, correct?
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So when the Suns gravity acts on Earth where is the opposite and equal force?
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I also have the same question for centripetal force in a planets orbit.
Newtonian Mechanics – Is Gravity a Force and What Is Its Opposite?
centripetal-forceforcesfree-body-diagramnewtonian-gravitynewtonian-mechanics
Related Solutions
They should feel the same. You only feel forces in orbit if there is something causing sensations, and nothing does in either case. Even on earth, you don't feel the "force" of gravity; you feel the force of the floor pushing you up so that you don't start falling under gravity's influence. In orbit, there is no floor, so you don't feel gravity.
You might imagine a force pushing on your spacesuit which then pushes against you. But in orbit, the force on your spacesuit causes an acceleration of the spacesuit that's exactly the acceleration that your body experiences, so there will be no relative difference in the motion of your body and your spacesuit, so you won't feel anything.
Or you might imagine your left arm feeling a different force than your right, and so feeling pulled apart. But the accelerations they experience in orbit will be the same, so their relative positions won't change, and you won't feel any difference.
(Technically, there might be minuscule differences referred to as tidal forces. These should be measurable by extremely sensitive instruments, but not by humans orbiting anything humans are likely to orbit any time soon.)
There is a misconception here.
The centripetal force is the force needed to make an object move in a circular path - or put differently, for the object to accelerate in a direction perpendicular to its motion.
When, for example, you swing a ball on a string in a circle, you provide that force through the tension in the string.
A car on a rollercoaster feels a force from gravity, and another force from the rails. When you are at the top of the curve, these two forces push in the same direction - and this means you will either:
- have less force from the rails (to the point where sometimes you are actually "hanging" from the rails - the wheels on a roller coaster grab from both the top and the bottom)
- go around a much tighter loop (also because you are typically going slower, you need the loop to be even tighter).
People are sometimes scared of the "virtual twin" of the centripetal force: the fictitious centrifugal force (which is very real in a rotating frame of reference). If you need the car to be kept on the rails by its motion at the top, it is sufficient that the centrifugal force ($F_c$, force pushing outwards) is greater than the force of gravity ($F_g$, pointing inwards). The net result will be an inward force by the rails ($F_r$); added to the force of gravity, this will keep the car on the rails.
Best Answer
As many others said, the Sun feels the same force towards Earth as the Earth feels towards the sun. That is your equal and opposite force. In practice though the "visible" effects of a force can be deduced through Newton's first law, i.e. ${\bf F} = m{\bf a}$. In other words, you need to divide the force by the mass of the body to determine the net effect on the body itself.
So:
${\bf F_s} = {\bf F_e}$
${\bf F_s} = m_s {\bf a_s}$
${\bf F_e} = m_e {\bf a_e}$
therefore,
$m_s {\bf a_s} = m_e {\bf a_e}$
and
${\bf a_s} = {\bf a_s} \frac{m_e}{m_s}$
Now, the last term is $3 \cdot 10^{-6}$! This means that the force that the Earth enacts on the sun is basically doing nothing to the sun.
Another way of seeing this:
$F = \frac{G m_s m_e}{r^2}$
$a_s = \frac{F}{m_s} = \frac{G m_e}{r^2}$
$a_e = \frac{F}{m_e} = \frac{G m_s}{r^2}$
$\frac{a_s}{a_e} = \frac{m_e}{m_s} = 3 \cdot 10^{-6}$
Again, the same big difference in effect.
Regarding the centripetal force, it is still the same force. Gravity provides a centripetal force which is what keeps Earth in orbit.
Note
It's worth pointing out that the mass that acts as the charge for gravity, known as gravitational mass is not, a priori, the same mass that appears in Newtons's law, known as inertial mass. On the other hand it is a fact of nature that they have the same value, and as such we may use a single symbol $m$, instead of two, $m_i$ and $m_g$. This is an underlying, unspoken assumption in the derivation above. This is known as the weak equivalence principle.