# 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 #2390768
2 months ago

Explanation:

It is known that the value of free fall acceleration is g = 6.44 . And, the value of radius of Earth is .

Let us assume that height of the satellite is h.

It is known that,

g =

where,   r = (R + h)

Hence,

r =

r =

=

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

h = r - R

=

=

Thus, we can conclude that the satellite is high above Earth's surface.

## Solution 2

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

where;

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

Thus;

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

h = 1.5 × 10⁶ m

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