4. A 1,750 kg weather satellite moves in a circular orbit with a gravitational

4. A 1,750 kg weather satellite moves in a circular orbit with a gravitational potential energy of 1.69 x 100 J. At its location, free-fall acceleration is only 6.44 m/s?. How high above Earth's surface is the satellite?

2 months ago

Solution 1

Guest Guest #2390768
2 months ago


It is known that the value of free fall acceleration is g = 6.44 m/s^{2}. And, the value of radius of Earth is 6.4 \times 10^{6} m.

Let us assume that height of the satellite is h.

It is known that,

             g = \frac{GM}{r^{2}}        

where,   r = (R + h)


              r = \sqrt{\frac{GM}{g}}

              r = \sqrt{\frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{6.44 m/s}}

                = 7.9 \times 10^{6} m

Now, formula for height of satellite above the Earth's surface is as follows.

            h = r - R

               = 7.9 \times 10^{6} - 6.4 \times 10^{6}

               = 1.5 \times 10^{6} m

Thus, we can conclude that the satellite is 1.5 \times 10^{6} m high above Earth's surface.

Solution 2

Guest Guest #2390769
2 months ago

The height of the satellite moving in a circular orbit above the Earth's surface is the satellite is; 1.5 × 10⁶ m

What is the height of gravitational fall?

The formula to find the height of the satellite above the earth's surface is; h = [√(GM/g)] - R


G is gravitational constant = 6.67 × 10⁻¹¹ N.m²/kg²

M is mass of earth = 5.98 × 10²⁴ kg

g is free fall acceleration = 6.44 m/s²

R is radius of earth  6400000 m


h = [√(6.67 × 10⁻¹¹ × 5.98 × 10²⁴/6.44)] - 6400000

h = 1.5 × 10⁶ m

Read more about height of gravitational fall at; brainly.com/question/14460830

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