Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Yesterday My friend asked me a question which put me into a confused state.

If a body of mass m is given a vertical thrust of $F$ such that $F > mg$, and the $F$ is allowed to remain applied on the body with constant acceleration, how high will the mass be lifted off ground?

He confused me asking that, as the body gains height, its potential energy should increase. So the body should stop going upward at some stage.

Though I said he is wrong, as energy is scalar and not vector, a doubt haunted me.

If I am able to give the force $F$ as said above constantly, will the body keep on going to outer space, as the weight $mg$ will never increase, but only decrease due to decrease in $g$ with altitude.

share|improve this question
    
Ask your friend why increasing potential energy would decelerate the body? When someone tells you such a statement, ask too see it written down in equation and you will find where the confusion lies. –  ja72 Jan 9 at 16:55
add comment

3 Answers

Like you said, the net force of the body keeps increasing, so the body does not have constant acceleration. However, keep in mind that the body will eventually approach the speed of light. That said, it will continue forever excluding external factors such as other planets, stars, etc.

share|improve this answer
add comment

Assuming that there are no forces other than gravity of Earth and the applied force (i.e. disregarding the effect of gravity due to other masses and drag), and using Newtonian mechanics, we can examine the scenario proposed by your friend with two pictures: (using $r$ to denote height of the mass)

1) Newton's Second Law

Since the a force of constant magnitude $F$ is applied, the net force acting on the mass is simply $F_n = F - mg$ away from Earth. Therefore, the equation of motion is $m \frac{dr}{dt} = F - mg$.

As you have rightly suggested, $g$ is not a constant, but varies according to Newton's Law of Gravitation $g(r) = \frac{GM}{r^2}$, which decreases as $r$ increases. Therefore, as the mass moves away from Earth, the net force acting on the mass increases, and with that the acceleration of the mass increases by Newton's second law.

There are practically no limits to how high the mass will go: mathematically from the equation of motion, $\lim_{t \rightarrow \infty} r(t) = \infty$ (if you decide to solve for the close form of $r(t)$); physically, you are essentially providing an acceleration away from Earth at every point of motion, there is practically no bound to which the mass will go.

2) Conservation of Energy

By providing a constant force $F$, you are, in energy terms, providing a work done contributing to the energy of the mass. The work done is given by $F \cdot r$, and this does not only provide energy to overcome the gravitational potential $-\frac{GMm}{r}$, but also the kinetic energy of the particle $\frac{1}{2} m \dot{r}^2$.

So part of the work done on the mass is used to provide energy for potential energy, while it also contributes to the increase in kinetic energy. As you can see here, as the mass moves further and further away from Earth, the gain in potential energy decreases (as it decrease with $\frac{1}{r}$), so higher proportion of the energy is contributed to the increase in kinetic energy, which matches the intuition we have in the force picture, where the acceleration increases as the mass moves away from Earth. Again, there are no bounds to how high the mass can travel.

share|improve this answer
add comment

If $F$ is constant, as long as you are close to the earth's surface, so will $F-mg$ be and the upward acceleration will be constant. As you get far from the earth, the gravitational attraction will decrease and the acceleration increase. You are continuing to add energy, so the increase in potential energy is no problem. It will keep going forever. After a while, the mass increase will slow the acceleration relative to earth, but it will still be positive.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.