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4. A small moon has radius 50 km. The acceleration due to gravity on the surface of the moon is 0.08 ms-2.

A small meteor of mass m is initially 200 km above the surface of the moon, with negligible velocity.

Assume that the only force on the meteor is the gravitational attraction of the moon, which is of magnitude mk / r^2 N (edited), where r is the distance of the meteor from the centre of the

moon, and k is a constant.

Find the value of k in S.I. units.

Find the speed of the meteor when it hits the surface of the moon.

A small meteor of mass m is initially 200 km above the surface of the moon, with negligible velocity.

Assume that the only force on the meteor is the gravitational attraction of the moon, which is of magnitude mk / r^2 N (edited), where r is the distance of the meteor from the centre of the

moon, and k is a constant.

Find the value of k in S.I. units.

Find the speed of the meteor when it hits the surface of the moon.

Last edited by Tsrnicolec; 3 weeks ago

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(Original post by

Have you had any thoughts / where are you stuck?

**mqb2766**)Have you had any thoughts / where are you stuck?

Last edited by Tsrnicolec; 3 weeks ago

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How I interpreted the question was that mk/r^2 = 0.08 and then thought initially it is of 250km (250,000m) away and when it hits the surface the distance is 50km (50,000m) away from the moon’s centre. But I’m not sure how k could be constant if the gravitational attraction and mass are the same, while r is different. Thanks for your reply! I really appreciate it How could we find out the value of k when there is more than one unknown?

**Tsrnicolec**)How I interpreted the question was that mk/r^2 = 0.08 and then thought initially it is of 250km (250,000m) away and when it hits the surface the distance is 50km (50,000m) away from the moon’s centre. But I’m not sure how k could be constant if the gravitational attraction and mass are the same, while r is different. Thanks for your reply! I really appreciate it How could we find out the value of k when there is more than one unknown?

You could use the ratio of the r^2 (surface and meteor) to get the acceleration of the meteor, which should get you k? Tbh, Im a bit confused where they say k is a constant. Where is the question from? Is there an "r^2" missing in the "mk" expression?

I suspect the k in the equation represents GM i.e. the mass of the moon and the gravitational constant. That would make the force

mk/r^2

(missing r^2 in the question) and r would be 50k or 250k and the acceleration is not constant as the inverse square law states.

Last edited by mqb2766; 3 weeks ago

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(Original post by

That wouldn't quite be right. Your right hand side (0.08) is an acceleration / what sort of a quantity is k?

You could use the ratio of the r^2 (surface and meteor) to get the acceleration of the meteor, which should get you k? Tbh, Im a bit confused where they say k is a constant. Where is the question from? Is there an "r^2" missing in the "mk" expression?

**mqb2766**)That wouldn't quite be right. Your right hand side (0.08) is an acceleration / what sort of a quantity is k?

You could use the ratio of the r^2 (surface and meteor) to get the acceleration of the meteor, which should get you k? Tbh, Im a bit confused where they say k is a constant. Where is the question from? Is there an "r^2" missing in the "mk" expression?

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It’s from integral maths for Cambridege International 9231 Further Mathematics.

**Tsrnicolec**)It’s from integral maths for Cambridege International 9231 Further Mathematics.

There is a missing r^2 in your OP, so

k = GM

which is constant and the acceleration is not constant (inverse square rule) and the force on the object is

mk/r^2 = ma

where a is the acceleration at a distance r from the centre of the moon (not constant).

Last edited by mqb2766; 3 weeks ago

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(Original post by

Edited ...

You could use the ratio of the r^2 (surface and meteor) to get the acceleration of the meteor, which should get you k? Tbh, Im a bit confused where they say k is a constant. Where is the question from? Is there an "r^2" missing in the "mk" expression?

I suspect the k in the equation represents GM i.e. the mass of the moon and the gravitational constant. That would make the force

mk/r^2

(missing r^2 in the question) and r would be 50k or 250k and the acceleration is not constant as the inverse square law states.

**mqb2766**)Edited ...

You could use the ratio of the r^2 (surface and meteor) to get the acceleration of the meteor, which should get you k? Tbh, Im a bit confused where they say k is a constant. Where is the question from? Is there an "r^2" missing in the "mk" expression?

I suspect the k in the equation represents GM i.e. the mass of the moon and the gravitational constant. That would make the force

mk/r^2

(missing r^2 in the question) and r would be 50k or 250k and the acceleration is not constant as the inverse square law states.

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#8

(Original post by

Oh sorry, yes there was the r^2 missing from the question. I’ve made changes on the thread accordingly. Thanks

**Tsrnicolec**)Oh sorry, yes there was the r^2 missing from the question. I’ve made changes on the thread accordingly. Thanks

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So are you ok now?

**mqb2766**)So are you ok now?

Last edited by Tsrnicolec; 3 weeks ago

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#10

What value did you get for k? Also note youre mixing m and km (not SI) quantities.

Edited ... When you do the integral, why not just use a definite integral between the two points (for r and v)? The lower limit is 250,000 and 0 and the upper limit is 50,000 and v.

Finally, if force is positive (increasing r) youd increase speed going further away. Force must be negative (point towards the centre) which I guess is the main problem here. Note the "magnitude" in the original question.

Edited ... When you do the integral, why not just use a definite integral between the two points (for r and v)? The lower limit is 250,000 and 0 and the upper limit is 50,000 and v.

Finally, if force is positive (increasing r) youd increase speed going further away. Force must be negative (point towards the centre) which I guess is the main problem here. Note the "magnitude" in the original question.

Last edited by mqb2766; 3 weeks ago

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