Question

25 km Core, 1.93 x 102 kg -Mantle, 4.01 x1024 kg Crust. 13490 km Figure 13-42 Problem 27.

Earth has a total mass of 5.98 x 10^24 kg and a radius of 6370 km Find the formula of a_g(r) in the core, that is, for 0 < r < 3490 km.

Find the formula of a_g(r) in the mentle, that is, for 3490 km < r < 6345 km.

Find the formula of a_g(r) in the crust, that is, for 6345 km < r < 6370 km.

Find the formula of a_g(r) outside the Earth, that is, for r > 6370 km.

Using the formulas obtain above, make a graph of gravitational acceleration, a_g(r) versus distance, r, from the center of Earth for 0 < r < 2.0R_E m, where R_E = 6370 km, the radius of Earth. [You must use a graph sheet and show the labels and scales with proper units for the axes used.]

Add the graph of a_g(r) when Earth is considered as a uniform sphere with the same total mass.

Answer 1

This is Basis axiss Theorem //g dx = uTG integral ((P dv ---(1) (1) CONSUMER SURFACE OF CORE MFA rho_core = mu_core/V_core = 1.93 times 10^24 kg/4/3 pi(3490 times 10^3)^3 m^3 P_core = 10844.58 kg/m^3 ---(2) g 4pi r^2 = -4pi 4 integral integral integral p 4pi r^2 dr g 4pi r^2 = -4pi 4 p 4pi r^3/3 g = -4pi 4r/3 g(r) - p_core4 pi 4 r/3 r for Delta < r < 3490 km G = 6.87 times 10^-11 P_core = 10844.58 kg/m^3 (ii) CONSUMER GAUUSSIAN OF RADIUS r IN 3490 km < r < 6345 km from eq (2) g 4pi r^2 = -4pi G[integral^3490_0 P_core dv + integral^r_3490 P_max dv] \r\n g 4pi r^2 = -4pi G[m_core + P_ integral^2_31 mkn dv] dv = 4pi r^2 dr { P_man = n_MAN/4/3 pi[(6345 times 10^3)^3 - (349.5 times 10^3)^3]m^3 = 4.04 times 10^24 kg/8.91 times 10^20 m^3 P_ = 4500.6 kg m^-3 g 4pi r^2 = -4pi G(M_core^-1 P_ 4/3 pi r^3 integral^2_3490 km) = -4pi G[m_cos + P_MAN 4pi/3[r^3 - (3490 times 10^3)^3)]