The period of the Earth and Mars rotating around its own axis is 0.997270 and 1.

Question

The period of the Earth and Mars rotating around its own axis is 0.997270 and 1.025956 days and 1.025956 days (1 day = 24 hours), respectively. The mass of the Earth and Mars is 5.9736 times 10^24 kg and 0.64185 times 10^24 kg, respectively. The gravitational acceleration on the surface of the Earth and Mars is 9.8 m/s^2 and 3.7 m/s^2, respectively. (a) Assume the shapes of the Earth and Mars are spherical, calculate the radius of the Mars. (b) Assume the shapes of the Earth and Mars are spherical, calculate the average densities for the Earth and Mars. (c) If we want to launch a geostationary satellite (a satellite whose orbit is such that it If remains above the same point on the planet's surface) orbiting around the Earth and Mars, planet (Earth or Mars) requires a higher altitude (i.e., the distance from the which surface of the planet?

Explanation / Answer

a) g_mars = G*M_mars/R_mars^2

==> R_mars^2 = G*M_mars/g_mars

R_mars = sqrt(G*M_mars/g_mars)

= sqrt(6.67*10^-11*0.64185*10^24/3.7)

= 3.40*10^6 m

similarly,

R_earth = sqrt(G*M_earth/g_earth)

= sqrt(6.67*10^-11*5.9736*10^24/9.8)

= 6.376*10^6 m

b)

density of earth = mass/volume

= 5.9736*10^24/(4/3*pi*(6.376*10^6)^3)

= 5501 kg/m^3

density of earth = mass/volume

= 0.64185*10^24/(4/3*pi*(3.40*10^6)^3)

= 3899 kg/m^3

c)

Time period of a satellite, T = 2*pi*(R + h)^(3/2)/sqrt(G*M)

as M is large h must be large.

so, Earth needs more altitude.

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