the near earth asteroid Rendezvous (NEAR), after travelling 2.1 billion km, is meant to orbit the asteroid Eros at a height of about 15 km from the asteroidal center. Eros is roughly box-shaped, with dimensions 40km x 6km x 6km. Assume eros has a density (mass/ volume) of about 2.3 x 10^3 kg/m^3.
a) what will be the period of NEAR as it orbits Eros?
b) If eros were a sphere witht he same mass and density, waht would be its radius.
c)what would g be at the surface of this spherical Eros?
This is an interesting problem. You will have to integrate the force of gravity from each dmass. Fortuantly, symettry makes it easier, as you can use symettry to just integrate one component of the force (the other side will cancel that component).
Because of symettry, make the box shape a wire of mass M and length 40km.
MassM is rho*volume.
So mass per unit length= rho*volume/40km
Now draw a triangle, one side the perpendiular of length 15km, the base of x (x goes from zero to 20km), and the hypotenuse of length=sqrt(225E3+x^2)
Now the angle theta at the apex of the triangle has a cosine of 15km/x. Notice that we can use the cosine of theta, as that is the force of attraction perpendicular to the rod.
Forcetotal=2*Gm INT dM/ sqrt(225E6+x^2) * cosTheta
but cosTheta= 15E3/sqrt(225E6+x^2)
and dM= rho*volume dx/40E3
so all that reduced is integrable.
Rember F/m= acceleartion which is equal to v^2/r and v= 2PIr/T so you can solve for T
Work it out for the sphere, just assume a point mass at 15km, see if it is different.
a) To find the period of NEAR as it orbits Eros, we can use the formula for the period of an orbiting object:
T = 2π√(r^3/GM)
Where:
T = Period of NEAR
r = Distance from the center of Eros to the orbiting NEAR (height + radius of Eros)
G = Gravitational constant (6.67430 × 10^-11 m^3 kg^-1 s^-2)
M = Mass of Eros
First, we calculate the distance from the center of Eros to the orbiting NEAR:
Distance = Height + Radius of Eros = 15 km + (40 km / 2) = 15 km + 20 km = 35 km = 35,000 m
Next, we calculate the mass of Eros:
Density = Mass / Volume
Volume = Length x Width x Height = 40 km x 6 km x 6 km = 960 km^3 = 960,000,000,000 m^3
Mass = Density x Volume = (2.3 x 10^3 kg/m^3) x (960,000,000,000 m^3) = 2.208 x 10^15 kg
Now, we can calculate the period of NEAR:
T = 2π√((35,000)^3 / (6.67430 × 10^-11) / (2.208 x 10^15))
T ≈ 4.9 hours
Therefore, the period of NEAR as it orbits Eros is approximately 4.9 hours.
b) If Eros were a sphere with the same mass and density, we can use the formula for the volume of a sphere to find its radius:
V = (4/3)πr^3
Where:
V = Volume of the sphere
r = Radius of the sphere
Since the volume and density remain the same (according to the given information), we can set up the following equation:
(4/3)πr^3 = 960,000,000,000 m^3
Now, we can solve for the radius:
r^3 = (3/4) x (960,000,000,000 m^3) / π
r ≈ 7755 m
Therefore, if Eros were a sphere with the same mass and density, its radius would be approximately 7755 meters.
c) To calculate the gravitational acceleration (g) at the surface of this spherical Eros, we can use the formula for gravitational acceleration:
g = (G x M) / r^2
Where:
g = Gravitational acceleration
G = Gravitational constant (6.67430 × 10^-11 m^3 kg^-1 s^-2)
M = Mass of Eros
r = Radius of Eros
Using the previously calculated mass and radius values:
g = (6.67430 × 10^-11 m^3 kg^-1 s^-2 x 2.208 x 10^15 kg) / (7755 m)^2
g ≈ 1.302 m/s^2
Therefore, the gravitational acceleration at the surface of this spherical Eros would be approximately 1.302 m/s^2.
To answer these questions, we will need to use some fundamental principles of physics and astronomy. Let's go step by step:
a) To find the period of NEAR as it orbits Eros, we can use Kepler's Third Law for gravitational orbits. Kepler's Third Law states that the square of the period of an orbit is proportional to the cube of the semi-major axis of the orbit. In this case, the semi-major axis is the distance between NEAR and the center of Eros, which is the sum of the radius of Eros and the orbit height (15 km).
First, let's convert the dimensions of Eros to meters:
Length (L) = 40 km = 40,000 m
Width (W) = 6 km = 6,000 m
Height (H) = 6 km = 6,000 m
Now, let's calculate the semi-major axis (a):
Semi-major axis (a) = Radius (r) + Orbit height (h)
a = (L/2 + W/2 + H/2) + h
a = (40,000/2 + 6,000/2 + 6,000/2) + 15,000
a = 46,500 m
Now, using Kepler's Third Law:
T^2 = k * a^3
T^2 = k * (46,500)^3
We need to find the value of k. For this, we can use the gravitational constant (G) and the mass (M) of Eros:
k = G * M
M = Density (ρ) * Volume (V)
Volume (V) = L * W * H
G = 6.67430 x 10^-11 m^3 kg^-1 s^-2
Given Density (ρ) = 2.3 x 10^3 kg/m^3:
V = (40,000) * (6,000) * (6,000)
V = 864 x 10^9 m^3
M = (2.3 x 10^3) * (864 x 10^9)
M = 1.9888 x 10^15 kg
Now, substitute the values of M and G into the equation to find k:
k = (6.67430 x 10^-11) * (1.9888 x 10^15)
k = 1.116 x 10^5 m^3 s^-2
Substituting the value of k into the equation for T^2:
T^2 = (1.116 x 10^5) * (46,500)^3
T^2 = 9.784 x 10^16
T = √(9.784 x 10^16)
T ≈ 312,605 seconds
Therefore, the period of NEAR as it orbits Eros is approximately 312,605 seconds.
b) If Eros were a sphere with the same mass and density, we can use the formula for the volume of a sphere to find its radius. The volume of a sphere (V) is given by the equation:
V = (4/3) * π * r^3
Since we know the density (ρ) and mass (M) of Eros, we can calculate the volume and then solve for the radius (r):
M = ρ * V
V = M / ρ
V = (2.3 x 10^3) * (864 x 10^9)
V = 1.9888 x 10^15
Now, use the formula for the volume of a sphere:
1.9888 x 10^15 = (4/3) * π * r^3
Solving for r:
r^3 = (3/4π) * 1.9888 x 10^15
r^3 ≈ 7.976 x 10^14
r ≈ ∛(7.976 x 10^14)
r ≈ 0.899 x 10^5
Therefore, if Eros were a sphere with the same mass and density, its radius would be approximately 0.899 x 10^5 meters.
c) To find the value of g at the surface of this spherical Eros, we can use Newton's law of gravitation. The formula for the acceleration due to gravity at the surface of a celestial body is:
g = G * M / R^2
Where G is the gravitational constant, M is the mass of Eros, and R is the radius of Eros.
Using the values calculated previously:
G = 6.67430 x 10^-11 m^3 kg^-1 s^-2
M = 1.9888 x 10^15 kg
R = 0.899 x 10^5 m
Substituting these values into the equation for g:
g = (6.67430 x 10^-11) * (1.9888 x 10^15) / (0.899 x 10^5)^2
Calculating this:
g ≈ 6.445 m/s^2
Therefore, the value of g at the surface of this spherical Eros would be approximately 6.445 m/s^2.