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2 years ago

How do i solve for the point of equigravity between two planets, given their mass ratio and the distance between them ?

1 answer:

4 0

The equigravity between two planets is determined from the product and their masses and square of distance between them.

<h3>Gravitational force between two planets</h3>

The gravitational force between two planets is calculated as follows;

F = GM₁M₂/R²

where;

G is universal gravitation constant
M₁ is mass of first planet
M₂ is mass of the second planet
R is the distance between the two planets

Use the mass ratio of the two planets to determine their individual masses.

Thus, the equigravity between two planets is determined from the product and their masses and square of distance between them.

Learn more about gravitational force here: brainly.com/question/72250

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OlgaM077 [116]

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3 years ago

The weight of an object is the product of its mass, m, and the acceleration of gravity, g (where g=9.8 m/s2). If an object’s mas

lina2011 [118]

You just told us that Weight = (mass) x (gravity),
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4 years ago

A woman (mass= 50.5 kg) jumps off of the ground, and comes back down to the ground at a velocity of -8.4 m/s.

Blizzard [7]

Answer:

Approximately $1.6\times 10^{3}\; \rm N$ .

Explanation:

By the Impulse-Momentum Theorem, the change in this woman's momentum will be equal to the impulse that is applied to her.

The momentum $p$ of an object is equal to the product of its mass $m$ and velocity $v$ . That is: $p = m \cdot v$ .

Let $v(\text{before})$ and $v(\text{after})$ represent the velocity of the woman before and after the landing. Let $m$ represent the woman's mass.

The woman's momentum before the landing would be $m \cdot v(\text{before})$ .
The woman's momentum after the landing would be $m \cdot v(\text{after})$ .

Therefore, the change in this woman's momentum would be:

$\begin{aligned}& \Delta p \\ & = p(\text{after}) - p(\text{before}) \\ &= m \cdot (v(\text{after})- v(\text{before}))\end{aligned}$ .

On the other hand, impulse is equal to force multiplied by the duration of the force. Let $F$ represent the average force on the woman. The impulse on her during the landing would be $F \cdot t$ .

Apply the Impulse-Momentum Theorem.

Impulse: $F\cdot t$ .
Change in momentum: $m \cdot (v(\text{after})- v(\text{before}))$ .

Impulse is equal to the change in momentum:

$F \cdot t = m \cdot (v(\text{after})- v(\text{before}))$ .

After landing, the woman comes to a stop. Her velocity would become zero. Therefore, $v(\text{after}) = 0\; \rm m \cdot s^{-1}$ .

$\begin{aligned}F &= \displaystyle \frac{m \cdot (v(\text{after})- v(\text{before}))}{t} \\ &= \frac{50.5\; \text{kg} \times \left(0 \; \mathrm{m \cdot s^{-1}}- 8.4\; \mathrm{m \cdot s^{-1}}\right)}{0.27\; \rm s} \\ &\approx 1.6 \times 10^{3}\; \rm N\end{aligned}$ .