# Oxford PAT 2014, Questions 11 and 12

## 8 thoughts on “Oxford PAT 2014, Questions 11 and 12”

1. jason says:

but if we consider this question in a bigger scale, for example, a galaxy. will the answer still be A? I am saying this because if we swing a mass attached to a rope on earth, the mass will fly along the tangent it the rope breaks. there are gravity from us on the mass as well, just insignificant.

1. Hi Jason, thanks for your comment.

The scenario in the question is that the ISS is orbiting the Earth, and a bit breaks off the ISS. The answer says that the bit that breaks off carries on orbiting the Earth with the ISS.

If, as in your example, you substitute a galaxy for the Earth and a solar system for the ISS, then you get the same answer: suppose an asteroid is pushed out of the solar system, it still continues to orbit the centre of the galaxy in the same way that the solar system does.

2. It means that we can assume that the ISS and the object that has detached from it are not moving relative to each other (i.e. when the object was detached it wasn’t shot out of a gun, it just drifted very slowly away).

3. James says:

What does the relative velocity of the ISS and the object is negligible mean?

4. Dan says:

Harry, are you preparing for 4th november exam?

5. Harry says:

1. Hi Harry, thanks for your question.

To understand the answer, think of a time T immediately after the object is detached from the ISS.

At T the positions of the ISS and the object are the same, as they have just that instant become detached.

The question tells us explicitly that the relative velocity of the ISS and the object is negligible after the object is detached, so the velocities are the same at T.

We can work out the acceleration of the ISS and the object by looking at the forces involved. At T the Earth’s gravity is acting on the ISS with a force proportional to the mass of the ISS, and acting on the object with a force proportional to the mass of the object. There are no other forces involved. Because force is mass times acceleration, and the forces on the ISS and the object are proportional to their respective masses, the accelerations of the ISS and the object are the same at T.

Since the position, velocity and acceleration are the same for the ISS and the object, they will follow the same path.

1. Harry says:

thanks