# Homework Help: Work due to varying acceleration

1. Mar 29, 2015

### Calpalned

1. The problem statement, all variables and given/known data
A 2800-kg space vehicle, initially at rest, falls vertically from a height of 3300km above the Earth's surface. Determine how much work is done by the force of gravity in bringing the vehicle to the Earth's surface.

**I read my textbook's method and I got confused. Please take a look at my questions at the bottom of this post.

2. Relevant equations
See part three.

3. The attempt at a solution
Here is my textbook's method.

I am confused:
1) Why do we have to use the negative vector form of the universal law of gravity?
2) The distance changes from 3300 km to 0 km (Earth's surface). All of the numbers between 0 and 3300 are positive, so is that why $d \vec l$ is positive?
3) Where did $d \vec l = dr \vec r$ come from?
4) During the integration $\vec r$ vanishes. Why is that?

Thank you.

2. Mar 29, 2015

### sk1105

1) We use the vector form because it includes the directions which are important in this question. The negative sign is there because the force of gravity acts in the opposite direction to $\vec r$.

2) $d\vec l$ is positive because it is change in $r$. As your textbook points out, you are moving from a large radius to a small radius, so the change will be positive.

3) Look more closely at the textbook. It says $d\vec l=dr\hat r$, where $\hat r$ means unit vector in the direction of $\vec r$. So $d\vec l$ is a vector of length $dr$ in the direction of $\vec r$.

4)Look at the integration. We have $\hat r \cdot (dr\hat r)$. Recall that the scalar product of two vectors will always produce a scalar, so there are no vectors in the final answer.

EDIT - Sorry I must have written answer (2) with my eyes closed! I think it's because you're looking at the work done by gravity, which is force over a distance, so this distance should be positive.

Last edited: Mar 29, 2015
3. Mar 31, 2015

### Calpalned

So whenever direction is important, I have to use the negative vector form of the universal law of gravity?

4. Mar 31, 2015

### Staff: Mentor

The real question is "why is $d \vec l = dr \vec r$ the differential displacement vector along the particle path?"

Notice that in the integration that they are doing, they are going from a larger radius to a smaller radius. So along the integration path, the lengths of the differential distances are (-dr). Note also that the direction of travel is in the negative r direction, as represented by $- \vec r$. So, the differential displacement vector along the particle path is $d \vec l = (-dr)(-\vec r)=dr \vec r$

Chet