A particle of mass \(m \mathrm {~kg}\), at a distance \(x \mathrm {~m}\) from the centre of the Earth, experiences a force of magnitude \(\frac { k m } { x ^ { 2 } } \mathrm {~N}\) towards the centre of the Earth, where \(k\) is a constant. Given that the radius of the Earth is \(6.37 \times 10 ^ { 6 } \mathrm {~m}\), and that a 3 kg mass experiences a force of 30 N at the surface of the Earth,
calculate the value of \(k\), stating the units of your answer.
The 3 kg mass falls from rest at a distance \(x = 12.74 \times 10 ^ { 6 } \mathrm {~m}\) from the centre of the Earth. Ignoring air resistance,
show that it reaches the surface of the Earth with speed \(7.98 \times 10 ^ { 3 } \mathrm {~ms} ^ { - 1 }\).
In a simplified model, the particle is assumed to fall with a constant acceleration \(10 \mathrm {~ms} ^ { - 2 }\). According to this model it attains the same speed as in (b), \(7.98 \times 10 ^ { 3 } \mathrm {~ms} ^ { - 1 }\), at a distance \(( 12 \cdot 74 - d ) \times 10 ^ { 6 } \mathrm {~m}\) from the centre of the Earth.