A satellite (pictured in brown) of mass 2400kg is in a circular orbit around the
ID: 1584969 • Letter: A
Question
A satellite (pictured in brown) of mass 2400kg is in a circular orbit around the earth at a distance 5R, from the earth's center (R, is the earth's radius which is about 6.37 x 106m Another satellite (pictured in grey) of the same mass as the first and traveling faster but going in the same direction, crashes into it as shown. After the crash, the two objects stick together. What velocity must the second grey satellite have, right before the collision, in order that the two objects end up escaping from the earth's gravitational field? (Don't forget that the acceleration due to gravity at r- Re is g-9.81 If you do the problem using symbols instead of numerical values as far as possible, you can use this information to simplify your equations.) km / sExplanation / Answer
In circular motion,
the centripetal force exerted on the satellite is balanced by the Gravitational Force by the Earth on the satellite.
i.e, mv2/r = GMm/r2 ---------> [1]
Where, m is the mass of the satellite = 2400 kg;
M is the mass of the Earth = 5.972*1024 kg;
G is the gravitational Constant = 6.67*10-11 N-m2/kg2
r is the distance of the satellite from the middle of the Earth = 5Re = 31.85*106 m
Re is the radius of the Earth = 6.37*106 m
v is the velocity of the satellite in its orbit.
from eqn[1],
v = sqrt [GM/r] = sqrt [6.67*10-11 N-m2/kg2*5.972*1024 kg/31.85*106 m]
vbrown, v = 3537.25 m/s (orbital velocity) [3.54 km/s]
By conservation of momentum, pi = pf
m*vbrown + m*vgray = (m+m)*ve -------->[2]
ve = escape velocity = sqrt[2]*orbital velocity
ve = 5002.43 m/s
from eqn[2],
m*[vbrown + vgray] = 2m*ve
vbrown + vgray = 2ve = 2*5002.43 m/s = 10 km/s
vgray = 10 km/s - vbrown = 10 km/s - 3.54 km/s = 6.46 km/s
the velocity of the gray satellite is vgray = 6.46 km/s
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