Welcome to Physics Problems Q&A, where you can ask questions and receive answers from other members of the community.

Prove that the vector of acceleration always points towards the origin

1 vote
139 views

A planet travels in an ellipse around the origin with the sun close to the origin. The planet's position in the ellipse is given by:

$x(t)=a\cos(kt)$

$y(t)=b\sin(kt)$

where $a>b$ and $k$ are constants and $0 ≤ t ≤ 2\pi$ ($t$ is a time variable).

The sun's position is given by $(x,y) = ( \sqrt{ a^2 − b^2}, 0)$.

I've learnt that taking the derivative of the functions for position twice gives the functions for the acceleration. How would this look and how can I compare this answer with the functions for positions to show that the vector of acceleration is always pointed towards the origin?

In my attempt I've got that the acceleration is

$$a_x(t) = −ak^2cos(kt), a_y(t) = −bk^2sin(kt)$$

So the difference between position and acceleration is $−k^2$, provided that I'm correct so far. However I do not understand how this would show that the acceleration always is pointed towards the origin. Does the place of the sun even have any relevance? Considering that no masses and gravitation is taken account for.

asked Dec 15, 2018 in Physics Problems by Pompan (110 points)
edited Dec 15, 2018 by sammy gerbil

1 Answer

0 votes

This is a very mixed-up question.

Planets move in elliptical orbits. which can be described by the parametric equations $$x(t)=a\cos(kt), y(t)=b\sin(kt)$$ However, in these equations in general t is NOT time. It is just a parameter with no physical meaning. These equations correctly describe the path ($x,y$) taken by the planet, but not the relations ($x, t$) and ($y, t$) between its position and time. That relation is much more complex, and cannot be described by any simple function.

The only case in which parameter $t$ represents time for a gravitational orbit, for which the force has the form $-K/r^2$, is when the orbit is a circle, ie when $a=b$. Then $k$ is the angular velocity. In general (ie when $a\ne b$) the angular velocity is not $k$.

The equations do give the correct relation between position and time for an orbit of a mass on a spring with zero natural length, ie a Hooke's Law central force which has the form $-Kr$. However, even for such an orbit $k$ is not the angular velocity, which is not constant - unless the orbit is circular.

How to show that the vector acceleration always points towards the origin

The functions you have obtained for acceleration are the components of a vector : $$a_x=-k^2x, a_y=-k^2y$$ The magnitude of this vector is $$a=\sqrt{a_x^2+a_y^2}=-k^2r$$ where $r$ is the distance of the object from the centre of the ellipse. This confirms that the central force is proportional to $r$ - ie it is a Hooke's Law force. You need to find the direction of this vector. The gradient of the vector wrt the x axis is $$m=\frac{a_y(t)}{a_x(t)}=\frac{y}{x}$$ This vector passes through the point $(x, y)$ and its gradient is $y/x$. From this information you can conclude that the force is always directed towards the origin which is at the centre of the ellipse.

Although the orbit is an ellipse, it has no relation to a gravitational orbit, for which the force is always directed towards one focus of the ellipse, where the Sun is located, not towards the centre of the ellipse.

answered Dec 15, 2018 by sammy gerbil (28,466 points)
edited Dec 30, 2018 by sammy gerbil
...