Gravitational force quick doubt

[rbnphlp]

Homework Statement

A spacecraft S of mass m moves in a straight line towards the centre of the earth. The earth
is modeled as a fixed sphere of radius R. When S is at a distance x from the centre of the
earth, the force exerted by the Earth on S is directed towards the centre of the Earth and has
magnitude k/x² , where k is a constant.
(a) Show that k = mgR^2.
(2)
Given that S starts from rest when its distance from the centre of the Earth is 2R, and that
air resistance can be ignored,
(b) find the speed of S as it crashes into the surface of the earth.

The Attempt at a Solution

I just have a qucick doubt on part B , I do get how to the question except the
markscheme states $a=-\frac{k}{x^2}$ where is the minus sign comming from , surely S is moving towards the Earth so isn't it in the same direction as the force?

Thanks

 

[kuruman]

Yes, the force is in the same direction as the acceleration. However, both force and acceleration are directed towards the the Earth, therefore both are negative. The convention is that positive is away from the Earth.

 

[rbnphlp]

Yes, the force is in the same direction as the acceleration. However, both force and acceleration are directed towards the the Earth, therefore both are negative. The convention is that positive is away from the Earth.

ok then would I be right in saying this :
since F=ma
$-m v\frac{dv}{dx}=-\frac{k}{x^2}$
and hence the negative cancel's out?
but the ms says:

$mv\frac{dv}{dx}=-\frac{k}{x^2}$

so I am not sure.

Thanks for the help so far

 

[kuruman]

Here v is the speed, a positive quantity. As the mass moves towards the centre, i.e. as x gets smaller, the speed increases and dv/dx is a negative ratio. On the other side k is positive and so is x². A negative sign is put in front of k/x² to ensure that the negative quantity on the left matches the negative quantity on the right.

 

[rbnphlp]

Here v is the speed, a positive quantity. As the mass moves towards the centre, i.e. as x gets smaller, the speed increases and dv/dx is a negative ratio. On the other side k is positive and so is x². A negative sign is put in front of k/x² to ensure that the negative quantity on the left matches the negative quantity on the right.

So if I have understood you right , youre agreeing my first equation is correct?

But the mark scheme of the paper says the 2nd one is correct? Is th emark scheme wrong?

thanks

 

[kuruman]

If v denotes the speed, then second equation is correct. The correct form of the first equation should be

$$mv \left|\frac{dv}{dx}\right|=\frac{k}{x^2}$$

Note that, because dv/dx is a negative quantity, both forms say the same thing.

 

[rbnphlp]

If v denotes the speed, then second equation is correct. The correct form of the first equation should be

$$mv \left|\frac{dv}{dx}\right|=\frac{k}{x^2}$$

Note that, because dv/dx is a negative quantity, both forms say the same thing.

Thank you very much