# Physics 1 course: inertial force in a rocketship [closed]

I am requesting a clarification to the solution of a question posed in my Physics 1 course.

Calculate the force acting on an astronaut with mass $$m$$ in a rocket that is starting on earth's surface with acceleration $$a$$.

The solution provided to this question states that the force is equal to $$-m(g(r)+a)$$, which makes sense to me. The next question then asks for the force after the rocket stopped its engine (so that $$a=0$$) (the flight occurs purely in the radial direction so it does not orbit the planet).

The solution to this second question then states that the inertial force on the astronaut is $$0$$, but given the equation from the first part there should still be a force equal to $$-m(g(r)+0) = -m g(r)$$ (since $$a=0$$, but $$g(r)$$ still acts on the ship).

Is the provided solution wrong or am I overseeing something?

• How does the "flight occur purely in the radial direction" as the earth is spinning and hence will impart it's spin to the rocket?!? – Mozibur Ullah Mar 18 at 4:54
• @MoziburUllah This was about a question that we got for our Physics 1 course so it does not have to be a 'realistic' scenario. Just replace 'Earth' by 'Planet that is not spinning' if you feel uncomfortable by the formulation of the question... – Lukas Schröder Mar 18 at 16:29
• Schoeder: Physics is about physical things and so questions on such are generally realistic - otherwise we're not doing physics but cartoon physics. Ensuring that the rotation of the earth is taken into account is part of a physics education ... – Mozibur Ullah Mar 19 at 21:16

Thanks to the OP for posting this question since it aims to crystallize the understanding of inertial forces versus non-inertial forces. The phenomenon described is that of a vehicle, of mass say $$M$$, with acceleration $$0 < a$$ initially and $$a = 0$$ later, which contains a passenger of mass $$m$$ that is at rest w.r.t. the vehicle at any time. The vehicle is traveling radially outward with radial position denoted by $$r$$ w.r.t. the earth which is assumed to be an inertial reference frame. We denote the (inertial) acceleration of the vehicle w.r.t. the earth by $$a_V := a$$, the (inertial) acceleration of the passenger w.r.t. the earth frame by $$a_P$$, the acceleration of the passenger, $$a_\text{PV}$$, as calculated (or observed) using a reference frame attached to vehicle, the thrust force acting on the vehicle by $$0 \geq F_V^\text{thrust}$$, the reaction contact force acted by the vehicle on the passenger by $$F_P^\text{react}$$, the external forces acting on the vehicle by $$F_{V}^\text{ext}$$ and the external forces acting on the passenger by $$F_{P}^\text{ext}$$. The goal is to calculate $$F_P^\text{react}$$.

The kinematics implies that since $$a_{PV} = 0$$, $$a_P = a_V + a_{PV}$$ implies that $$a_P = a_V = a$$ and the dynamics using the law of motion implies that $$F_V^\text{ext} = F_V^\text{thrust} - M g(r) = M a_V = M a$$ and $$F_P^\text{ext} = F_P^\text{react} - m g(r) = m a_{P} = m a$$. These two equations are summarized as $$F_V^\text{thrust} - M g(r) = Ma,$$ $$F_P^\text{react} - m g(r) = ma.$$ Notice that in general, the equation of motion $$m a_{PV} = m a_P - m a_V = m a_P - F_V^\text{ext}$$ implies that the non-inertial forces on $$P$$ due to the calculation of the acceleration w.r.t. the non-inertial (in general) reference frame $$V$$ are given by $$- m a_V = - m a$$.

1. In the first situation, since $$0 < a$$, $$F_V^\text{thrust} = M(g(r) + a)$$ and $$F_P^\text{react} = m(g(r) + a)$$.
2. In the second situation, since $$F_V^\text{thrust} = 0$$, $$a = -g(r)$$ so that $$F_P^\text{react} = 0$$.
• Thanks for the detailed answer :) – Lukas Schröder Mar 13 at 21:30
• @LukasSchröder you're welcome. Such discussions appear trivial but require subtle understanding of Newton's laws and can therefore become confusing. – kbakshi314 Mar 13 at 21:31