# Can you solve this difficult physics question?

A planet of unknown mass has a radius of (4.960x10^3) km. If a wheel from a spacecraft travelling with a velocity of (1.500x10^1) m/s down falls from and strikes the ground (5.871x10^4) m below in (8.0x10^1) s, what is the mass of the planet?

Thank you soooo much! This helped me understand. But, I am confused on one part, how did you know it was to the power of 24?

### 2 Answers

- electron1Lv 71 year agoFavorite Answer
When an object is on the surface of a planet, the weight of the object is equal to the Universal gravitational force.

Weight = m * g

Fg = G * M * m ÷ r^2

m * g = G * M * m ÷ r^2

g = G * M ÷ r^2

G = 6.67 * 10^-11

r = 4.96 * 10^6 meters

Let’s use the following equation to determine the acceleration.

d = vi * t + ½ * a * t^2

5.871 * 10^4 = 15 * 80 + ½ * a * 80^2

3200 * a = 57,510

a = 57,510 ÷ 3200

This is approximately 18 m/s^2. This is value of g.

(57,510 ÷ 3200) = 6.67 * 10^-11 * M ÷ (4.96 * 10^6)^2

M = (57,510 ÷ 3200) * 2.46016 * 10^13 ÷ 6.67 * 10^-11

The mass of this planet is approximately 6.63 * 10^24 kg. I hope this is helpful for you.

- oubaasLv 71 year ago
"travelling" is meaningless in physics ...so 3 possible cases are investigated :

a)

spacecraft is moving away from the planet

-58710 = 15*80-g1/2*80^2

g1 = (58710+15*80)*2/80^2 = 18.72 m/sec^2

b)

spacecraft is moving towards the planet

-58710 = -15*80-g2/2*80^2

g2 = (58710-15*80)*2/80^2 = 17.97 m/sec^2

c)

spacecraft is orbiting around the planet

-58710 = -g3/2*80^2

g3 = 58710*2/80^2 = 18.35 m/sec^2

since g = M*6.673*10^-11/(4.96^2*10^12) = 2.71*10^-24*M

then :

Ma = 18.72*10^24/2.71 = 6.91*10^24 kg

Mb = 17.97*10^24/2.71 = 6.63*10^24 kg

Mc = 18.35*10^24/2.71 = 6.77*10^24 kg

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I UNDERSTAND EVERYTHING, just how did you get to the power of 13?

So (18 * 4.960*10^3km)^2 = 442828800

This is where I am confused