2. is a variable that gives location relative to an origi

A satellite in a circular orbit of radius R around planet X has an orbital period T. If Planet X had one-fourth as much mass, th

Iteru [2.4K]

<h2>Answer: 2T</h2>

According to the Third Kepler’s Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $R$ of its orbit.

This Law is originally expressed as follows (in the case of planet X and assuming we have a circular orbit):

$T^{2}=\frac{4\pi^{2}}{GM}R^{3}$ (1)

Where:

$G$ is the Gravitational Constant

$M=1.9(10)^{27}kg$ is the mass of planet X

$R$ is the radius of the orbit of the satellite around planet X

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=2\pi\sqrt{\frac{R^{3}}{GM}}$ (2)

Now, we are asked to find the period when tha mass of the planet is $\frac{1}{4}M$ . In order to do this, we have to rewrite equation (2) with this new value:

$T=2\pi\sqrt{\frac{R^{3}}{G(\frac{1}{4}M)}}$ (3)

Solving:

$T=4\pi\sqrt{\frac{R^{3}}{G(\frac{1}{4}M)}}$ (4)

On the other hand, if we multiply both sides of equation (2) by 2, we have:

$2T=4\pi\sqrt{\frac{R^{3}}{GM}}$ (5)

As we can see, (5) is equal to (4). This means the orbital period is twice the orignal period.

Hence, the answer is:

If Planet X had one-fourth as much mass, the orbital period of this satellite in an orbit of the same radius would be 2T.

3 0

3 years ago

What's the answers for all of the questions ?///

Maru [420]

More energy, colour and I think less bright

8 0

3 years ago

A boy drops a ball from an observation tower. The ball hits the ground in 5.0 s. What is the ball's velocity at the time of impa

Orlov [11]

The answer is C. 49 m/s

The kinetic equation is:
v2 = v1 + a * t

v1 - initial velocity
v2 - final velocity
a - gravitational acceleration
t - time

We know:
v2 = ?
v1 = 0 (in free fall
a = 9.8 m/s
t = 5

v2 = v1 + a * t
v2 = 0 + 9.8 * 5
v2 = 0 + 49
v2 = 49 m/s

7 0

3 years ago

PLEASEEEEEE HEEEEEEEEEEEEEEELPPPPPPPPPPPPPPPPP I NED HELP WHIT THIS!!!!! D:

aleksley [76]

The meters per second
+1t a second / 2t

6 0

2 years ago

one of the four wave interactions is very important to how plants use light to grow. guess which interaction this is

KiRa [710]

The photosynthetic wave interaction between visible light and a photosensitive part of a plant is very important t how plants use light to grow.
The frequency range and intensity levels of this light, I don't know.

Maybe the above ???

6 0

2 years ago