The gravitational field strength at the surface of a planet of radius R is g . A satelliteis moving in a circular orbit a distance R above the surface of the planet. What is themagnitude of the acceleration of the satellite
Question
The gravitational field strength at the surface of a planet of radius R is g . A satelliteis moving in a circular orbit a distance R above the surface of the planet. What is themagnitude of the acceleration of the satellite
Solution
The acceleration of the satellite is given by the formula for gravitational acceleration:
a = GM/(R+r)^2
where:
- G is the gravitational constant,
- M is the mass of the planet,
- R is the radius of the planet, and
- r is the distance of the satellite from the surface of the planet.
Given that the gravitational field strength at the surface of the planet (r=0) is g, we can write:
g = GM/R^2
We can solve this equation for M:
M = gR^2/G
Substituting this into the equation for a gives:
a = G(gR^2/G)/(R+R)^2 = gR^2/(2R)^2 = g/4
So the magnitude of the acceleration of the satellite is g/4.
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