A 11.0 kg satellite has a circular orbit with a period of 1.80 h and a radius of 7.50 × 106 m around a planet of unknown mass. I
1 answer:
Answer:
Explanation:
Expression for times period of a satellite can be given as follows
Time period T = 1.8 x 60 x 60
= 6480
T² = where T is time period , r is radius of orbit , G is gravitational constant and M is mass of the satellite.
6480² = 4 x 3.14² x 7.5³ x 10¹⁸ / GM
GM = 4 x 3.14² x 7.5³ x 10¹⁸ / 6480²
= 3.96 X 10¹⁴
Expression for acceleration due to gravity
g = GM / R² where R is radius of satellite
20 = 3.96 X 10¹⁴ / R²
R² = 3.96 X 10¹⁴ / 20
= 1.98 x 10¹³ m
R= 4.45 x 10⁶ m
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Answer:
m = 28.7[kg]
Explanation:
To solve this problem we must use the definition of kinetic energy, which can be calculated by means of the following equation.
where:
Ek = kinetic energy = 1800 [J]
m = mass [kg]
v = 11.2 [m/s]
Answer:
density = 5520 kg/m^3
Explanation:
given that
radius of earth = 6378 km
G = 6.67 x 10⁻¹¹ m³/kg.s²
g = 9.80 m/s²
we know,
mass of earth
M = 5.972 x 10²⁴ kg
density =
V = volume of the earth = 4/3πr³
V = 4/3 x 3.14 x (6378 x 10³)³
V = 1.08 x 10²¹ m³
density =
density = 5.52 x 10³ kg/m^3
density = 5520 kg/m^3