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?
Any help would me much appreciated
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?
Any help would me much appreciated

Explanation / Answer

Radius R = 3000 km = 3 * 10 ^ 6 m height h = 92.5 m time taken t= 5 s we know from the relation h = ut + ( 1/ 2) gt^ 2                                          h = ( 1/ 2) g t ^ 2   Since initial speed u = 0                                         g = 2h / t^ 2                                            = 7.4 m / s ^ 2 we know g = GM / R ^ 2 from this mass of the planet M = gR ^ 2 / G where G = Gravitational constant = 6.67 * 10 ^ -11 N m ^2 / kg ^ 2 plug the values weget M = 9.985 * 10 ^ 23 kg (a). density of the planet D = mass / volume                                          = M / ( 4/ 3) R ^ 3                                         = 3M / 4R ^ 3                                         = 0.08828 * 10 ^ 5 kg / m ^ 3                                         = 8828 kg / m ^ 3 (b).  the escape velocity from that planet v =[ GM / R ]                                                                 =4.711 * 10 ^ 3 m / s (c). Initial velocity v = 10 m / s timetaken t = [ 2h / g ]                   = [ 2* 92.5 / 7.4] = 5 s Horizontal distance R = v t                                  = 10 m / s * 5 s= 50 m      

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---