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 Answer

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  • 1 decade ago
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    Part 1

    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.

    Part 2

    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.

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