# Math Help - Work-kinetic energy theorem

1. ## Work-kinetic energy theorem

The magnitude of the attractive force of gravity between two bodies is F = GMm/r 2. G is a constant equal to 6.67×10−11 N·m2/kg2, M and m are the masses, and r is the distance between the centers of the two bodies. The gravitational force of a star of mass 1.14×1032 kg and radius 3.48×109 m is the sole force acting on a rocket of mass 1.25×106 kg. The rocket is stationary relative to the star at distance of 1.85×1011 m. Sadly, the rocket has exhausted its fuel, and it will be pulled to its doom inside the star. How fast will it be moving when it reaches the surface of the star?

2. Originally Posted by Linnus
The magnitude of the attractive force of gravity between two bodies is F = GMm/r 2. G is a constant equal to 6.67×10−11 N·m2/kg2, M and m are the masses, and r is the distance between the centers of the two bodies. The gravitational force of a star of mass 1.14×1032 kg and radius 3.48×109 m is the sole force acting on a rocket of mass 1.25×106 kg. The rocket is stationary relative to the star at distance of 1.85×1011 m. Sadly, the rocket has exhausted its fuel, and it will be pulled to its doom inside the star. How fast will it be moving when it reaches the surface of the star?
$G = 6,67 \times 10^{-11}$

$m_1 = 1,14 \times 10^{32}$

$m_2 = 1,25 \times 10^6$

$r = 1,85 \times 10^{11}$

I would suggest you work out $F$, and then maybe you can use $F = ma$ to determine the acceleration, but phisycs isn't my field. This is a Topsquark question...

3. Originally Posted by Linnus
The magnitude of the attractive force of gravity between two bodies is F = GMm/r 2. G is a constant equal to 6.67×10−11 N·m2/kg2, M and m are the masses, and r is the distance between the centers of the two bodies. The gravitational force of a star of mass 1.14×1032 kg and radius 3.48×109 m is the sole force acting on a rocket of mass 1.25×106 kg. The rocket is stationary relative to the star at distance of 1.85×1011 m. Sadly, the rocket has exhausted its fuel, and it will be pulled to its doom inside the star. How fast will it be moving when it reaches the surface of the star?
Originally Posted by janvdl
This is a Topsquark question...
I honor your respect of me.

It probably could be done janvdl's way, but the problem is that the acceleration is not constant, so Newton's 2nd will be difficult to apply.

I like the suggestion that Linnus put in the title: use the Work-Energy Theorem.

Gravity is a conservative force, so
$\Delta E = 0$

Some definitions here. I'm going to set the center of the star as the 0 point for the gravitational potential energy. (Yes, we can get away with that. At least, I was shocked when I first saw that.) Positive r will be "up," or in the outward radial direction from this point.

So
$\Delta PE + \Delta KE = 0$

$\left ( -\frac{GmM}{r} - -\frac{GmM}{r_0} \right ) + \left ( \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 \right ) = 0$
(where m is the mass of the rocket, and M is the mass of the star.)

Since the rocket starts from rest:
$\left ( -\frac{GmM}{r} + \frac{GmM}{r_0} \right ) - \frac{1}{2}mv^2 = 0$

Thus:
$v = \sqrt{ \frac{2}{m} \left ( \frac{GmM}{r} - \frac{GmM}{r_0} \right )}$

$v = \sqrt{ \frac{2GM}{r} - \frac{2GM}{r_0} }$

Now, $r_0 = 1.85 \times 10^11~m$ and $r = 3.48 \times 10^9~m$. For reference I get a solution of $v = 2.0707 \times 10^6~m/s$.

-Dan