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adell [148]
2 years ago
9

With a bit of algebraic reasoning find your gravitational acceleration toward any planet of mass M a distance d from its center.

Physics
1 answer:
grandymaker [24]2 years ago
5 0

The acceleration due to gravity is given as:

                             g = GM/r²

<h3>Derivation of gravitational acceleration:</h3>

According to Newton's second law of motion,

F = ma

where,

F = force

m = mass

a = acceleration

According to Newton's law of gravity,

F<em>g </em>= GMm/(r + h)²

F<em>g = </em>gravitational force

From Newton's second law of motion,

F<em>g </em>= ma

a = F<em>g</em>/m

We can refer to "a" as "g"

a = g = GMm/(m)(r + h)²

g = GM/(r + h)²

When the object is on or close to the surface, the value of g is constant and height has no considerable impact. Hence, it can be written as,

g = GM/r²

Learn more about gravitational acceleration here:

brainly.com/question/2142879

#SPJ4

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Two equal, but oppositely charged particles are attracted to each other electrically. The size of the force of attraction is 48.
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Given:

The force of attraction is F = 48.1 N

The separation between the charges is

\begin{gathered} r=\text{ 60.9 cm} \\ =\text{ 60.9}\times10^{-2}\text{ m} \end{gathered}

Also, the magnitude of charge q1 = q2 = q.

To find the magnitude of charge.

Explanation:

The magnitude of charge can be calculated by the formula

\begin{gathered} F=\frac{k(2q)}{r^2} \\ q=\frac{Fr^2}{2k} \end{gathered}

Here, k is the Coulomb's constant whose value is

k\text{ = 9}\times10^9\text{ N m}^2\text{ /C}^2

On substituting the values, the magnitude of charge will be

\begin{gathered} q=\frac{48.1\times(60.9^\times10^{-2})^2}{2\times9\times10^{^9}} \\ =9.91\times10^{-10}\text{ C} \\ =9.91\times10^{-4}\text{ }\mu C \end{gathered}

Thus, the magnitude of each charge is 9.91 x 10^(-4) micro Coulombs.

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