Posted by Rebecca on Thursday, December 27, 2012 at 9:00am.
F = 9.8 N = 1*Me * G /(6.4*10^6)^2
so
Me = 9.8 *41*10^12 / 6.67*10^-11
= 60.2 * 10^23 kg
Ve = (4/3) pi Re^3 = (4/3)(pi)(6.4*10^6)^3
= 1098*10^18
Rho = mass/volume = (6.02/1.098)10^(24-21)
= 6020/1.098 = 5483 kg/m^3
about 5.5 times density of water, lots of molten metal and heavy rock in there.
density ρ =M/V
mg=W=GmM/R² =>
M= WR²/Gm.
V=4πR³/3.
ρ =3 WR²/4πR³Gm =
=3W/4πRGm =
=3•9.8/4•3.14•6.4•10⁶•6.67•10⁻¹¹•1=
=5.48•10³ kg/m³
Related Questions
physics-mechanics - A 1.0 kg mass weighs 9.8 N on Earth's surface, and the ...
PHYSICS - Search: a 1.0 kg mass weighs 9.8 n on earths surface and the radius of...
Physics - Using the fact that a 1.0 kg mass weighs 9.8N on the surface of the ...
Physics - A 1.0kg mass weighs 9.8 Newton on earth's surface, and the radius ...
physics - the mass of a body is 50 kg on the surface of the earth.find its ...
physics - A 3.3 kg mass weighs 30.03 N on the surface of a planet similar to ...
physics - A 2.5 kg mass weighs 23.75 N on the surface of a planet similar to ...
physics - A 2.5 kg mass weighs 23.75 N on the surface of a planet similar to ...
earth - A particle weighs 120 N on the surface of the earth. At what height ...
Physics please help - How far from the center of the Earth (in km) is the center...
For Further Reading