A free fall experiment is performed on a newly discovered spherical planetoid (\

Question

A free fall experiment is performed on a newly discovered spherical planetoid ("planet x"), where an 80 kg astronaut dropped a rock from a height of 216 cm and its falling time was electronically measured to be 1.2 sec. The planet's circumference was determined to be 36.74 x 10^6 m by geometrical methods. The astronaut is fully aware that h = 1/2gt^2. Determine the following: A) The weight of the astronaut on the planet's surface: B)The mass M of planet X: C) the orbital speed V(orb) near planet X's surface: D) The orbital period T near planet X's surface: E)The altitude h for a geo-synchronous satellite orbiting planet X with a rotational period of 15 hrs:

Explanation / Answer

A free fall experiment is performed on a newly discovered spherical planteroid (named planet Z). where an astronaut dropped a rock from a height of 288 cm and its falling time was electronically measures to be 1.2 seconds. the planets circumfrence was determined to be 36.74 * 10^6m by geometrical methods. The ausronaut is fully aware that h=1/2gt^2
find
1. the acceleration of graviy g at the planets surface
2. The mass M of planet Z
3. The orbital speed Vorb near the planets surface
4) the orbital period T near the planet's surface
5) the altitude h for a geosynchronous satellite orbiting planet Z with rotational period of 18.2 hrs

• ANS

A free fall experiment is performed on a newly discovered spherical planteroid (named planet Z). where an astronaut dropped a rock from a height of 288 cm and its falling time was electronically measures to be 1.2 seconds. the planets circumfrence was determined to be 36.74 * 10^6m by geometrical methods. The ausronaut is fully aware that h=1/2gt^2
find
1. the acceleration of graviy g at the planets surface
2. The mass M of planet Z
3. The orbital speed Vorb near the planets surface
4) the orbital period T near the planet's surface
5) the altitude h for a geosynchronous satellite orbiting planet Z with rotational period of 18.2 hrs

• ANS

A free fall experiment is performed on a newly discovered spherical planteroid (named planet Z). where an astronaut dropped a rock from a height of 288 cm and its falling time was electronically measures to be 1.2 seconds. the planets circumfrence was determined to be 36.74 * 10^6m by geometrical methods. The ausronaut is fully aware that h=1/2gt^2
find
1. the acceleration of graviy g at the planets surface
2. The mass M of planet Z
3. The orbital speed Vorb near the planets surface
4) the orbital period T near the planet's surface
5) the altitude h for a geosynchronous satellite orbiting planet Z with rotational period of 18.2 hrs I'm going to do the maths here with some idealized values for ease of calculation. Your answers should be nearly the same, but a little different on the decimals.

So, on my planet, the astronaut drops the rock from a height of 300 cm = 3 m, it falls for 1s, and the planet's radius is 6 * 10^6 m, so that its circumference is C = 2r = 37.7 * 10^6 m.

1. Solving the equation you have for g gives:

g = 2 h / t² = 2 * (3m) / (1 s)² = 6 m/s².

2. Since g = G M/r² where G = 6.673 * 10^{-11} m³/(kg s²), we have that:

M = g r² / G = 6 m/s² * (6 * 10^6 m)² / (6.673e-11 m³/(kg s²))
M = 216 * 10^12 / (6.673 * 10^{-11}) kg = 32.4 * 10^23 kg = 3.24 * 10^24 kg.

3. In circular motion, r = r, v = r, a = ² r, where = 2 / T. Solving for a in terms of v gives:

a = v² / r

Which means that v = ( r a ) = ( 6 * 10^6 m * 6 m/s² ) = 6 000 m/s.

4. You can use either the above expression for a or for v to find and thus T. Probably my favorite phrasing is this, due to its intuitive simplicity:

T = 2 / = 2 r / v = C / v.

T = 37.7 * 10^6 m / (6000 m/s) = 6 283 s.

This is a bit under two hours, as an hour is 3600 s.

5. Now we want to combine the above expressions:

GM / r² = ² r.

With a little manipulation, this can just as easily be:

r³ = k T²

...for some complicated constant k, which must be given by our existing numbers:

k = r³ / T² = (6 * 10^6 m)³ / (6 283 s)² = 5.47 * 10^12 m³ / s²

Thus for 18.2 h = 65 520 s, I get:

r³ = (65 520 s)² * (5.47 * 10^12 m³ / s²) = 2.35 * 10^22 m³

Taking the cube root:

r = 2.86 * 10^7 m.

Its altitude above ground is therefore 2.86 * 10^7 m - 6 * 10^6 m = 2.26 * 10^7 m. I'm going to do the maths here with some idealized values for ease of calculation. Your answers should be nearly the same, but a little different on the decimals.

So, on my planet, the astronaut drops the rock from a height of 300 cm = 3 m, it falls for 1s, and the planet's radius is 6 * 10^6 m, so that its circumference is C = 2r = 37.7 * 10^6 m.

1. Solving the equation you have for g gives:

g = 2 h / t² = 2 * (3m) / (1 s)² = 6 m/s².

2. Since g = G M/r² where G = 6.673 * 10^{-11} m³/(kg s²), we have that:

M = g r² / G = 6 m/s² * (6 * 10^6 m)² / (6.673e-11 m³/(kg s²))
M = 216 * 10^12 / (6.673 * 10^{-11}) kg = 32.4 * 10^23 kg = 3.24 * 10^24 kg.

3. In circular motion, r = r, v = r, a = ² r, where = 2 / T. Solving for a in terms of v gives:

a = v² / r

Which means that v = ( r a ) = ( 6 * 10^6 m * 6 m/s² ) = 6 000 m/s.

4. You can use either the above expression for a or for v to find and thus T. Probably my favorite phrasing is this, due to its intuitive simplicity:

T = 2 / = 2 r / v = C / v.

T = 37.7 * 10^6 m / (6000 m/s) = 6 283 s.

This is a bit under two hours, as an hour is 3600 s.

5. Now we want to combine the above expressions:

GM / r² = ² r.

With a little manipulation, this can just as easily be:

r³ = k T²

...for some complicated constant k, which must be given by our existing numbers:

k = r³ / T² = (6 * 10^6 m)³ / (6 283 s)² = 5.47 * 10^12 m³ / s²

Thus for 18.2 h = 65 520 s, I get:

r³ = (65 520 s)² * (5.47 * 10^12 m³ / s²) = 2.35 * 10^22 m³

Taking the cube root:

r = 2.86 * 10^7 m.

Its altitude above ground is therefore 2.86 * 10^7 m - 6 * 10^6 m = 2.26 * 10^7 m. I'm going to do the maths here with some idealized values for ease of calculation. Your answers should be nearly the same, but a little different on the decimals.

So, on my planet, the astronaut drops the rock from a height of 300 cm = 3 m, it falls for 1s, and the planet's radius is 6 * 10^6 m, so that its circumference is C = 2r = 37.7 * 10^6 m.

1. Solving the equation you have for g gives:

g = 2 h / t² = 2 * (3m) / (1 s)² = 6 m/s².

2. Since g = G M/r² where G = 6.673 * 10^{-11} m³/(kg s²), we have that:

M = g r² / G = 6 m/s² * (6 * 10^6 m)² / (6.673e-11 m³/(kg s²))
M = 216 * 10^12 / (6.673 * 10^{-11}) kg = 32.4 * 10^23 kg = 3.24 * 10^24 kg.

3. In circular motion, r = r, v = r, a = ² r, where = 2 / T. Solving for a in terms of v gives:

a = v² / r

Which means that v = ( r a ) = ( 6 * 10^6 m * 6 m/s² ) = 6 000 m/s.

4. You can use either the above expression for a or for v to find and thus T. Probably my favorite phrasing is this, due to its intuitive simplicity:

T = 2 / = 2 r / v = C / v.

T = 37.7 * 10^6 m / (6000 m/s) = 6 283 s.

This is a bit under two hours, as an hour is 3600 s.

5. Now we want to combine the above expressions:

GM / r² = ² r.

With a little manipulation, this can just as easily be:

r³ = k T²

...for some complicated constant k, which must be given by our existing numbers:

k = r³ / T² = (6 * 10^6 m)³ / (6 283 s)² = 5.47 * 10^12 m³ / s²

Thus for 18.2 h = 65 520 s, I get:

r³ = (65 520 s)² * (5.47 * 10^12 m³ / s²) = 2.35 * 10^22 m³

Taking the cube root:

r = 2.86 * 10^7 m.

Its altitude above ground is therefore 2.86 * 10^7 m - 6 * 10^6 m = 2.26 * 10^7 m.

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