# Physics......?

a) Find the gravitational field strength of an asteroid with the mass of 3.2 * 10^3 kg and an average radius of 30 km when at a distance of 3 km from its surface

b) if an astronaut popped out of a worm hole (at rest) at 3 km from the asteroid

i) how long would it take him to fall to the asteroid’s surface?

ii) how fast would he be traveling when he hit it? (Assume acceleration stays constant)

iii) if a year is 3.16 * 10^7 s, how many years would it take for the astronaut to reach the asteroid?

Please show work, thank you

### 2 Answers

- billrussell42Lv 71 month agoBest Answer
Gravitational attraction in newtons

F = G m₁m₂/r²

G = 6.674e-11 m³/kgs²

m₁ and m₂ are the masses of the two objects in kg

r is the distance in meters between their centers

g = G m/r² = (6.674e-11)(3200) / (33000)² = 1.961e-16 m/s²

falling object starting from rest

h is height in meters, t is time falling in seconds,

g is acceleration of gravity, usually 9.8 m/s²

v is velocity in m/s

h = ½gt²

t = √(2h/g)

v = √(2gh)

simple solution, assuming g does not change

t = √(2•3000/1.961e-16 m/s²) = 5.53e9 s

v = √(2gh) = √(2•1.961e-16•3000) = 1.08e-6 m/s

5.53e9 s x 1 yr/3.16e7s = 175 yrs

- oubaasLv 71 month ago
a)

g = M*G/((3.3*10^4))^2 = 3.2*10^3*6.67*10^-11*10^-8/3.3^2 = 2.0*10^-16 m/sec^2

b)

3*10^2 = a/2*t^2

t = √3*10^2/10^-16 = 10^9√3 sec

c)

V = a*t = 2*10^-16*10^9√3 = 3.4*10^-7 m/sec

d)

t = 10^9√3 / 3.16*10^7 = 55 years