An astronaut lands on a new planet of radius R= 3000 km, and wantsto measure its

Question

An astronaut lands on a new planet of radius R= 3000 km, and wantsto measure its density. In order to do that, he drops a stone ofunknown mass from the top of the space shuttle, which has a heightof 92.5m and measures the time it takes it to reach the ground. Hemeasures t = 5s.
a) What is the density of the planet? b) What is the escape velocity from that planet? c) If the astronaut, would throw the stone from the top of theshuttle horizontally with an initial velocity 10m/s, how long wouldit take the stone to reach the ground? And how far from the shuttlewould it land?
a) What is the density of the planet? b) What is the escape velocity from that planet? c) If the astronaut, would throw the stone from the top of theshuttle horizontally with an initial velocity 10m/s, how long wouldit take the stone to reach the ground? And how far from the shuttlewould it land?

Explanation / Answer

We'll assume that the acceleration of gravity is relativelyconstant for the range that the stone free-falls to the planet'ssurface. We can derive a second degree equation byintegrating the following equation twice WRT time: ay =-g (the minus sign is because the force of gravity isdownward). The equation we get is: y = -(1/2)*g*t2+ v0*t + y0; where g is the acceleration ofgravity, v0 is initial vertical velocity, y0is the initial height, and t is time. We know everythingexcept g, so plug in the values and solve for g: 0 = -(1/2)*g*(5s)2 + (0)*(5s) + 92.5m g = (92.5m)/((1/2)*(5s)2) = 7.4m/s2 To compute the planet's mass we need Newton's law of gravitationand his second law of motion: F = m*a = G*m*Mp/r2 a = G*Mp/r2 Mp = a*r2/G =(7.4m/s2)*((3000km)*(1000m/km))2/(6.673*10-11m3/kg-s2)= 9.98*1023kg [The (1000m/km) term converts kilometers to meters.] We now need to compute the volume of the planet: Volume = (4/3)**r3 =(4/3)*(3.14159)*((3000km)*(1000m/km))3 =1.131*1020m3 a) The density is: = mass/volume =(9.98*1023kg)/(1.131*1020m3) =8824kg/m3 Normally, density is in grams per cubic centimeter, so let'sconvert: (8824kg/m3)*(1000g/kg)/(100cm/m)3 =8.824g/cm3 b) Escape velocity can be computed by integrating Newton'sacceleration formula from the surface to infinity; however, we'lljust use the well know formula for escape velocity: ve = (2*G*Mp/r)0.5 ve =(2*(6.673*10-11m3/kg-s2)*(9.98*1023kg)/((3000km)*(1000m/km)))0.5= 6663m/s c) Throwing a stone horizontally doesn't change the verticaltime-distance to fall (assuming we aren't throwing the stone farenough for the curvature of the planet to have some effect). The stone will take the same time: 5 seconds. The distanceis: d = v*t = (10m/s)*(5s) = 50m.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.