Question

Question: A very early, simple satellite consisted of an inflated spherical aluminum balloon 30m in diameter and of mass 20 kg . Suppose a meteor having a mass of 7.0 kg passes within3.0 mof the surface of the satellite. What is the magnitude of the gravitational force on the meteor from the satellite at the closest approach?

Short answer

Answer:

The magnitude of the gravitational force on the meteor from the satellite is$F=2.9\times {10}^{-11}\text{N}$

Step-by-step solution

Identification of given data

The mass of the spherical aluminum balloon is M = 20 kg

The mass of the meteor is m = 7.0kg

The diameter of the spherical aluminum balloon is d = 30 m

The distance between the meteor and the surface of a satellite is x = 3.0 m

Significance of Newton’s law of universal gravitation

Every particle in the universe is attracted to every other particle with a force that is directly proportional to the product of the masses and inversely proportional to the square of the distance, according to Newton's Law of Universal Gravitation.

Formula:

$F=\frac{GMm}{r^2}$

Where, is the gravitational constant ( $6.67\times {10}^{-11} \frac{N.m^2}{k{g}^2}$)

Determining the magnitude of the gravitational force on the meteor from the satellite at the closest approach

The radius of the spherical aluminum balloon is

$$\begin{gathered} R=\frac{d}{2} \\ =\frac{30\text{m}}{2} \end{gathered}$$

R = 15

The distance betweenthecenter ofthesatellite and the center ofthemeteor is

$r=R+x$

According to the Newton’s law of gravitation, the force of attraction is

$$\begin{gathered} r=15\text{m}+3.0\text{m} \\ r=18\text{m} \end{gathered}$$

The magnitude of the gravitational force on the meteor from the satellite is

$$\begin{gathered} F=\frac{GMm}{r^2} \\ F=\frac{\left(6.67\times {10}^{-11}\frac{{\text{N.m}}^{\text{2}}}{{\text{kg}}^{\text{2}}}\right)\times \left(20\text{kg}\right)\times \left(7.0\text{kg}\right)}{{\left(18\text{m}\right)}^2} \\ F=2.9\times {10}^{-11}\text{N} \end{gathered}$$

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