Physics, 13.11.2019 00:31, jaustin1961

Asatellite has a mass of 372 kg and is located at 2.01 ✕ 106 m above the surface of earth. (a) what is the potential energy associated with the satellite at this location? your response differs from the correct answer by more than 10%. double check your calculations. j (b) what is the magnitude of the gravitational force on the satellite?

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Answers

Gravitational potential energy of the satellite: approximately $(\rm -1.77 \times 10^{10}\; J)$ . (Note that this value is negative.)

Magnitude of the gravitational force between the earth and the satellite: approximately $\rm 2.11\times 10^{3}\; N$

Explanation:

(a)

Consider the formula for the gravitational potential energy (GPE) between two dot or spherical mass:

$\displaystyle \text{GPE} = -\frac{G \cdot m \cdot M}{r}$ ,

where

$G \approx \rm 6.67 \times 10^{-11}\; N \cdot m^{2} \cdot kg^{-2}$ is the gravitational constant,

and

are the mass of the two objects, and

is the separation between the center of mass of the two objects.

Note the negative sign in front of the fraction. Electrostatic force might either be attractive or repulsive. However, gravity between two object of mass is always attractive. As a result, unlike electrical potential energy, the gravitational potential energy between two objects with mass should always be negative.

Keep in mind that the center of mass of a sphere with uniform density is at its center. If the earth is approximated as a sphere like that, its center of mass wouldn't be on the surface (i.e. the ground.) Rather, the center of the earth's mass should be somewhere deep underground near its core.

Look up the following quantities:

Mean radius of the earth (average distance between the core and the surface:) approximately $\rm 6.371 \times 10^{6}\; m$ .Mass of the earth: approximately $\rm 5.972 \times 10^{24}\; kg$ .

The distance between the satellite and the center of mass of the earth comes in two parts:

distance between the satellite and the surface of the earth, anddistance between the surface of the earth and its center of mass.

In other words,

$r = \rm \underbrace{\rm 2.01 \times 10^{6}\; m}_{\text{Satellite and ground}} + \underbrace{\rm 5.972 \times 10^{6}\; m}_{\text{Ground and Center of Mass}} \approx 7.98 \times 10^{6}\; m$ .

Make sure that all values are in standard units and apply the formula:

$\begin{aligned}\text{GPE} &= -\frac{G \cdot m \cdot M}{r} \\ &= - \frac{6.67 \times 10^{-11}\times 372 \times 5.972 \times 10^{24}}{6.371 \times 10^{6}} \\ & \rm \approx -1.77 \times 10^{10}\; J\end{aligned}$ .

Since all inputs are in their standard units, the unit of the output (potential energy) should be in joules, which is the standard unit of energy.

(b)

The formula for the magnitude of gravitational force between two points (or sphere) of mass is

$\displaystyle \text{Gravitational Force} = \frac{G \cdot m \cdot M}{r}^{2}$ .

Note that as a magnitude, the output of this formula should always be positive. Evaluate this formula for the same , , , and as in part (a) of this problem:

$\begin{aligned}\text{Gravitational Force}&= \frac{G \cdot m \cdot M}{r^{2}} \\ &= \frac{6.67 \times 10^{-11}\times 372 \times 5.972 \times 10^{24}}{\left(6.371 \times 10^{6}\right)^{2}} \\ & \rm \approx 2.11 \times 10^{3}\; N\end{aligned}$ .