Applying Kepler's Laws


The distance of a m=9.00×106 kg m = 9.00 \times 10 ^6 \text{ kg} planet from the Sun is R=2.00×109 km. R = 2.00 \times 10^9 \text{ km}. If we stopped the planet in orbit and then let it fall straight towards the Sun, approximately how long would it take for the planet to reach the Sun in seconds?

Consider both the planet and the Sun as point masses.
The mass of the Sun is M=2.00×1030 kg. M = 2.00 \times 10^{30} \text{ kg}.
Gravitational constant is G=6.67×1011 Nm2/kg2. G = 6.67 \times 10^{-11} \text{ Nm}^2\text{/kg}^2.

A satellite, moving in an elliptical orbit, is 520 km 520 \text{ km} above Earth's surface at its farthest point and 260 km 260 \text{ km} above at its closest point. Calculate the semimajor axis of the orbit.

The radius of Earth is Re=6.37×106 m. R_e = 6.37 \times 10^6 \text{ m}.

A 5×104 kg 5 \times 10^4 \text{ kg} geo-stationary satellite revolves around the earth in a circular orbit of radius 36000 km. 36000 \text{ km}. Approximately what will be the time period of a 5×104 kg 5 \times 10^4 \text{ kg} satellite in an orbit of radius 12000 km? 12000 \text{ km}?

A geo-stationary satellite is a satellite whose position in the sky remains the same in the view of a stationary observer on earth.

If the distance between the earth and the sun were 9.0% 9.0 \% of its present value, the approximate number of days in a year would be _______.

A satellite is in an elliptical orbit. If the maximum orbital velocity of the satellite is vM=2.60vm, v_M = 2.60 v_m, where vmv_m is the minimum orbital velocity, what is the approximate orbital eccentricity?


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