physics help..whats the min speed needed for stone to escape asteroid?
An astronaut is standing on a spherical asteroid of mass 7.5 × 10^17 kg and
radius 41 km.
1)The astronaut wonders if he can throw a stone sufficiently fast for it to
escape the gravitational pull of the asteroid and travel on into outer
space. Derive an expression for the minimum speed the stone would
have to have to escape from the asteroid and show that its value is
49 m s^−1.
2)The astronaut cannot throw the stone this fast. He can only manage
35 m s^−1. He thinks, however, that he might just be able to put the
stone into orbit around the asteroid if he throws it horizontally. Show
that he is correct.
can any one show me how its done step by step please.
- 1 decade agoFavorite Answer
If we start with the Law of Gravitation
F_g = GMm_2 / r^2
where G = 6.67 x 10^-11 N.m^2.kg^-2
M = mass of the planet
m = mass of the body (stone)
r = radius (in this case = 41km)
We next look at the gravitational potential energy.
U = - GMm / r
If the kinetic energy of the stone thrown from the planet were equal in magnitude to the potential energy, it could escape from the planet.
1/2 mv^2 = U = -GMm/r
Therefore the escape velocity is:
v = √(2GM/r)
Putting in the values we get:
v = √(2 x 6.67e-11 x 7.5e17 / 41000)
v = 49.4 ms^-1.
Assuming a circular orbit, the centripetal acceleration of a satellite in circular orbit has an acceleration with magnitude = v^2/r and is always directed towards the centre of its orbit.
Equating the centripetal force to the gravitational force:
mv^2/r = GMm/r^2
We get the equation for the speed required to keep the stone in orbit:
v = √GM/r
Input the values:
v = √6.67e-11 x 7.5e17 / 41000
v = 34.9 ms^-1
Hope this helps and makes sense.Source(s): http://hyperphysics.phy-astr.gsu.edu/hbase/vesc.ht... http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.ht... http://hyperphysics.phy-astr.gsu.edu/Hbase/cf.html