Law of Universal Gravitation?

Question

I got a question:

You are an astronaut who has been sent to a large rocky asteroid which is essentially spherical in shape. You are standing on the surface of the asteroid and you need to make an estimate of its radius, R_a. You know that the density (rho)
(density is the mass per unit volume of a substance) of the
asteroid is 2.80 x 10^3 kg m^-3 and that this density is uniform throughout.
You have equipment that enables you to measure the time it takes for a small object to fall from rest through a distance of 1.00m. Your measurement is 4.20s

(a) calculate the acceleration due to gravity g_a at the surface of the asteroid.
You may assume that g_a is constant over the one metre drop in the
experiment.

Can I use the equations of uniform motion for this one? Or do I have to use
density, along with the law of universal gravitation?

(b) Obtain an expression for g_a in terms of the gravitational constant G, the
radius if the asteroid R_a and the density (rho) of the asteroid. Hence
determine R_a. (The volume of a sphere of Radius R is given by 4/3 pi R^3)

Thanks.

Hello everyone
I was able to do the next part about escape speed(v_esc = 180.56 ms^-1). But I'm stuck on the next question:

You decide that your time would be better spent checking your measurement of g_a
using a different method. You have in your equipment a standard 1 kg mass and accurate spring balance. You know that on Earth at a place with g = 9.81 ms^-2
that a mass of 10.00g produced an extension of 50.0 mm. On the asteroid surface
you suspend your standard kilogram mass from the spring and nmeasure an
extension of 57.6 mm. Does this confirm your first measurement of g_a.

Cheers.

Answer 1

Yes, in part (a) you can use the eq of uniform motion.
(a) s = ut + 1/2gt^2
1 = 0 + 1/2g(4.2)^2
1 = 8.82g
g = 0.113 m/s^2

(b) Here you have to use the density
weight = gravitational force
mg = GMm/r^2...... cancel m on both sides and instead of M use density x volume
g = G x 2.8 x 10^3 x 4/3 x 3.14 x R
g = 1.172 x 10^4 GR


now to find R all you have to do is substitute the value of g from part (a) and the value of the constant G in this eq .

Answer 2

(a) To find g, use h = ut + 1/2 gt^2; since u = 0, you have g = 2h/t^2, where t = 4.2 sec and h = 1 m.

(b) F = mg = GmM/r^2; so that g = GM/r^2 and r^2 = GM/g. M = rho V = rho 4/3 pi r^3; where rho is the density and V the volume of the asteroid. Thus, r^2 = Grho 4/3 pi r^3/g and 1 = Grho(4/3)pi r/g; so that g = (4/3) G rho pi r and 3g/(4 G rho pi) = r = R_a.

Lesson learned, put down what you know about the problem. Write whatever relationships and equations correspond to what you know. Check them out, if you have everything known and the problem is workable, a lot of things will cancel out or have substitutable factors...like M = rho 4/3 pi r^3.

Answer 3

on the century while Newton lived, anybody believed there replaced into one set of regulations for products right here on earth, and a distinctive set of regulations for celestial bodies. Newton got here to comprehend that his regulation of gravitation utilized in the two domain names, as a result the be conscious prevalent replaced into extra to the call, to tension this ingredient.

Answer 4

Maybe it's an oscillation question?