Astronomers launch a satellite to explore a nearby asteroid that has come dangerously close to the earth.The satellite begins orbit around the earth,with an orbital radius 6.85 X 10^3 Km, before being moved to an orbit about the asteroid with a period of 9.06 X 10^3 S at a distance of 311 m from its center.
A) what is the speed of the satellite as it orbits Earth(who's mass is 5.9736 X 10^24 Kg)?
B)what is the mass of the asteroid?
A)when in orbit about earth
mv^2/r = m (G Mearth)/r^2
solve for v
B)
same equation but get v immediately
v T = 2 pi r
then
m v^2/r = m Ma(G/r^2)
m , mass of sat cancels
we know v
solve for Ma
Im having troubles putting it in so A would equal 767km/h?!!?! and B would equal 345 KG!?
v^2 = G Me/r
v^2 =
(6.67*10^-11)(5.97*10^24)
/6.85*10^6
v^2 = 5.81 * 10^7 = 58.1*10^6
so
v = 7.62*10^3
so
v = 7620 meters/second
Thats B only? im so lost be more clear please
B)
v(9.06*10^3) = 2 pi (311)
v = .216 m/s
I do not think so, suspect you mean 311km not 311 m
that would give v = 216 m/s
then
7620^2 = G Ma/r
7620^2 = 6.67*10^-11 Ma/311
Ma = 2.71*10^20 kg
A is 7620 m/s
B is 2.71*10^20 kg
and for heaven's sake use SCI units, do not use km and hours or something. Use meters, seconds, Newtons, kilograms.
1 hour = 3600 seconds
1 km = 1000 meters
So what would the mass of the asteroid be...?
READ
A is 7620 m/s
B is 2.71*10^20 kg