Question

A satellite orbits Earth. The only force on the satellite is the gravitational force exerted by Earth. How does the satellite’s acceleration compared to the gravitational field at the location of the satellite?

Answer 1

Answer:

here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location

Explanation:

As we know that gravitational field is defined as the force experienced by the satellite per unit of mass

so we will have

$E = \frac{F}{m}$

now in order to find the acceleration of the satellite we know by Newton's II law

$F = ma$

so we will have

$a = \frac{F}{m}$

so here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location

Answer 2

Given that $m <<< M$, then $a_{m}$ is considerably greater than $a_{M}$, which tends to be zero. ($a_{M}\to 0$).

Let be $m$ and $M$ the masses of the satellite and the Earth, respectively. By the Newton's Law of Gravitation, the accelerations experimented by the satellite and the Earth ($a_{m}$, $a_{M}$), in meters per square second, are, respectively:

Satellite

$a_{m} = \frac{G\cdot M}{R^{2}}$ (1)

Earth

$a_{M} = \frac{G\cdot m}{R^{2}}$ (2)

Where:

• $G$ - Gravitation constant.
• $R$ - Distance between the centers of the Earth and the satellite.

If we know that $m <<< M$, then $a_{m}$ is considerably greater than $a_{M}$, which tends to be zero. ($a_{M}\to 0$).

We kindly invite to check this question on Newton's law of gravitation: https://brainly.com/question/24023449