Ask question

Ask question

All categories

2 years ago

7

Astronomers have discovered a new planet called "Xandar" beyond the orbit of Pluto (No, not really but I need a fake planet for

this problem!). It has a satellite called "Jim" that orbits with a period of 12.9 days at a distance of 72,600 km. What is the mass of "Xandar"?

1 answer:

Burka [1]2 years ago

3 0

Answer:

m = 1.82E+23 kg

Explanation:

G = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

mG = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

m = 1.82E+23 kg

You might be interested in

Find an expression for the acceleration a of the red block after it is released. use mr for the mass of the red block, mg for th

Drupady [299]

<span>Assuming pulley is frictionless. Let the tension be ‘T’. See equation below.</span>

<span> </span>

6 0

1 year ago

Read 2 more answers

During the 440, a runner changes his speed as he comes out of the curve onto the home stretch from 18 ft/sec to 38 ft/sec over a

Sloan [31]

Answer:

$6.67ft/s^2$

Explanation:

We are given that

Initial velocity=u=18ft/s

Final velocity,v=38ft/s

Time=t=3 s

We have to find the average acceleration over that 3 s period.

We know that

Average acceleration,a= $\frac{v-u}{t}{t}$

Using the formula

Average acceleration,a= $\frac{38-18}{3}ft/s^2$

Average acceleration,a= $\frac{20}{3}ft/s^2$

Average acceleration,a= $6.67ft/s^2$

Hence, the average acceleration= $6.67ft/s^2$

5 0

1 year ago

Taylor places a nail on a bar magnet. The nail sticks to the magnet when lifted up off the table. She touches a paperclip to the

Ratling [72]

Prior to touching the bar magnet, the magnetic domains in the nail were pointing in random directions. When Taylor touched the nail to the bar magnet the magnetic fields of the magnetic domains aligned and it became a temporary magnet.

5 0

1 year ago

Read 2 more answers

Rod A has twice the diameter of rod B, but both are made of iron and have the same initial length. Both rods are now subjected t

Vilka [71]

Answer:

ΔLa/ΔLb = 1

Explanation:

The change in length of a solid is given by the following formula:

ΔL = α L ΔT

where,

ΔL = Change in length

α = coefficient of linear expansion

L = Original Length

ΔT = Change in Temperature

Since, the length and change in temperature for both rods are same. Also, the material of each rod is same, which implies that coefficient of linear expansion for both rods is same. Hence, the ratio of change in length of both rods will be:

<u>ΔLa/ΔLb = 1</u>

4 0

2 years ago

Calculate the mass of the air contained in a room that measures 2.50 m x 5.50 m x 3.00 m if the density of air is 1.29 g/dm3.53.

Law Incorporation [45]

Answer:

$5.32\cdot 10^4 g$

Explanation:

First of all, we need to find the volume of the room, which is given by

$V=2.50 m \cdot 5.50 m \cdot 3.00 m =41.3 m^3$

Now we can find the mass of the air by using

$m=dV$

where

$d=1.29 g/dm^3$ is the density of the air

$V=41.3 m^3 = 41,300 dm^3$ is the volume of the room

Substituting,

$m=(1.29)(41300)=5.32\cdot 10^4 g$

6 0

1 year ago