physics questions about gravity. pleasse help?

Question

1. an astronaut lands on a planet that has twice the diameter and mass of earth. how does the astronaut weight differ from that of earth.

2.if the earth were uniform density [same mass and volumes throughout ] what would the value of g be inside the earth at half its radius.

3.the mass of a certain neutron stae is 3.0 x 10^30 kg [1.5 solar masses] and its radius is 8000 m [8 km]. what is the acceleratiin of gravity at the surface of this burnt out condensed star

Answer 1

we have

g= GM/R^2

1) So if M'= 2M and R'= 2R then g'= g/2
2) Gravity decreases with height, at a rate which near the surface of the Earth is such that linear extrapolation would give zero gravity at a height of one half the radius of the Earth, i.e. the rate is 9.8 m·s−2 per 3200 km.


Alternatively, we can consider a spherically symmetrical Earth and subtract from the mass of the Earth that of the shell outside the point of observation, because that does not cause gravity inside. This gives the same result.


3) apply g=GM/R^2

Answer 2

The formula of gravitation's force is F = km m' /r^2 where k
constant m' mass of the body and m the mass of the planet (or star), r is the distance to the center of planet (or star)

1) if M increases by 2 but the distance increases by 2 as the distance is squared in the denominator the force willbe divded by 2, and he will weight the half.

2) if the distance to the center is divide by two. the denominaator will be divided by 4 and g would be 4g (4 times more)

3) I have found mass of the earth 6*10^24kg. Compared to the mass of the star we have

mass of the star /mass of earth is 5*10^5 so m' =5 10^5 m of
earth the radius of earth is 800 times the radius of the star

so the new gravity wiil be g *5 10^5/(1/800)^2 = 5 10^5 *6.4 10^5 =3.2 10^11 g

Answer 3

1) Weight is defined as W=mg where m is the mass of the body and g is the acceleration due to gravity of the other body attracting mass m. In general g=GM/r^2 where G is Newton's gravitational constant, M is the mass of the attracting body (earth or planet) and r is the radius of the assumed spherical body. To find out how much an astronaut would weigh it is sufficient to compare g on the 2 planets...

For the planet g = 2GM(earth)/[4r(earth)]^2 = GM(earth)/8[r(earth)]^2. Answer: the astronaut would weigh 8 times less or 1/8 of his weight on earth.

2) You need some easy calculus to answer this one:

Assuming that at generical distance r from the center of the earth the value of g(r) is costant on the sphere of radius r, we can express the elementary mass dm realtive to the spherical cap at distance r as dm= ρdV (as density ρ is constant) where dV is the volume element of the infinitesimal spherical cap dr thick. So...

dm=ρdV=ρ(4πr^2)dr where 4πr^2 is the area of the sphere with radius r. Substituting this expression for the elementary mass dm in g's expression (g=GM/r^2) we get..

dg=Gdm/r^2 =Gρ(4πr^2)dr/r^2 and integrating over half the sphere (earth) to answer the question we get...

g(inside, at R/2)=4/3 GρπR/2 which is half the value of g at the earth's surface as g(R)=4/3 GρπR...

3)As usual g=GM/R^2 = (6.6742 10^-11 m^3/s^2*kg) * (3.0 x 10^30 kg) / (6.372797 m)^2 = 4.9 10^18 m/s^2 which is roughly 10^18 times earth's value!!!! Probably a neutron star or a black hole!

Answer 4

1. g=G*m/r^2 (g being surface gravitational acceleration, m being planet mass), so double r and m and you cut g in half.
2. The total g would be the sum of the g from the shell of matter outside the measurement point plus the g from the solid sphere inside that radius. From Newton's shell theorem discussed in the 2 references, the shell's g=0. (Ref. 1 is a detailed derivation, ref. 2 is brief and more intuitive.) Then you have the inside sphere's g, with half the radius and 1/8 the mass (.5^3=1/8). The result is g is half the surface value.
3. Just substitute m and r in g=G*m/r^2. (The value of G is not known to a high degree of precision. Typical values quoted are 6.672E-11 to 6.675E-11.)

Answer 5

1 the astronaut weighs half on the planet

g=GM/Rsq

Therefore g(planet)/g(earth) = M(p)xRsq e divided by R sq planet x mass earth

g'/g=2mR/4Rm

g'=g x 1/2

g' on planet is half of that of the earth

weight = mg' =mg/2

therefore weight planet = 1/2 weight earth

2 the value of g would be one fourth of that in the surface

Answer 6

So, u pretty much use the same formula
since F=mg=w (weight) and F=G*M*m/R2 (R is squared), where G- is the gravitational constant, M is the mass of the planet and R is its radius and m is the weight of austronaut (assume 70kg), you assume astronaut's weight because it is irrelevant
1. F=G*M*m/R2=
=(6.67E-11)*(2*5.98E24)*70/(2*6.37E6)2
=344N

2. g=G*M/R2=
=(6.67E-11)*(5.98E24)/(0.5*6.37E6)=
=39.32 m/s2

3. g=6.67E-11*(3E30)/(8E3)2=
=3.1265E12 m/s2 as you would expect
Hope that will help.

Answer 7

i think as long as we are imagining issues that defy actuality, we ought to easily think of it may artwork. nonetheless identity be intrested in the form you may get in or out of the spheres with out slicing a hollow in them, which might lead them to non-uniform, or we ought to easily think of ourselves interior. on the grounds that we in all probability stepping into some long journeys can i think of myself a cheeseburger? What appropriate to the gasoline? is this machinge powered by making use of fairy airborne dirt and mud or mind's eye?