Question

In empty space, an object appears 70cm away from a 9.8kg mass. What is its acceleration towards the mass?

Answer 1

is there a time.. or any other info?

Answer 2

is there any more info?
anyways here are 4 equations that deal with acceleration:
a = f/m
and
a = change in v/ change in time
and
s = s +vt+(1/2)at^2
and
v^2 = v^2 + 2as

Answer 3

Empty space means there are only those two masses. the mass that appears will only feel the gravitational force of the body M = 9.8 kg. Gravitational force is\[F = g \frac{m M}{r^2}\]
M is the body's mass, r = 0.7 m is the distance between the objects, g is the gravitational constant.
F is accelerating the object. \[ F = m a = g \frac{m M}{r^2}\]
\[=> a = g \frac{M}{r^2}\]

Answer 4

Is this right?\[a=\frac{6.67\times 10^{-11}N\cdot m^2/kg^2\times 9.8kg}{(0.70m)^2}=1.3\times 10^{-9}m/s^2\]Is this right?

Answer 5

I would have though that you also have to think about the mass of the "object" in your question. Best wishes, Alistair

Answer 6

If the mass doubles, the force accelerating the particle also doubles. The acceleration stays the same.

Can someone please confirm \(1.3\times 10^{-9}\)?

Answer 7

I've also got \(a = 1.3 \times 10^{-9} ms^{-2}\)

Answer 8

Thank you kindly.