The Earth may be assumed to be an isolated sphere of radius 6.4 × 103 km with its mass of 6.0 × 1024 kg concentrated at its centre. An object is projected vertically from the surface of the Earth so that it reaches an altitude of 1.3 × 104km.(adsbygoogle = window.adsbygoogle || []).push({});

Calculate, for this object:

(i) the change in gravitational potential

(ii) the speed of projection from the Earth’s surface, assuming air resistance is

negligible.

My Attempt

i) I applied the formula

φ = - GM/radius of earth + altitude of object

= GM (1/radius of earth - 1/altitude of object)

= 3.17 x 10^10 J/kg

Don't know what I'm doing wrong here, this is not the correct answer.

The marking scheme of this past paper states the following solution:

change = 6.67 x 10^-11 x 6.0 x 10^24 x({6.4 x 10^6}-1 - {1.94 x 10^7}-1)

change = 4.19 x 107 J/kg

I am not sure how they calculated the bold part?

ii)

I used the Escape Velocity Formula as follows:

Vesc = squareroot{[2GM/r]} or squareroot[2gr]

Vesc = 3.54 x 10^5

Again this is not correct answer. So when do we use this escape velocity formula exactly if not in this case?

Again the marking scheme has the following suggested solution:

½mv2 = mdeltaφ

v2 = 2 x 4.19 x 107 = 8.38 x 107

v = 9150 m s-1

Here I'd like to ask, how come we have equated gravitational potential and kinetic energy? Aren't they two different things?

Thanks in advance!

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# Change in Gravitational Potential

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