Gravitational pull on asteroid/Period it takes for one orbit

• testme
In summary: I'm not sure where the discrepancy might be. However, in summary, to find the time it takes for an asteroid to orbit the Earth at an altitude of 2800km, you can use the formula t = 2πr/v, where r is the sum of the radius of the Earth (6.37 x 10^6 m) and the altitude of the asteroid (2800 km), and v is the velocity calculated using the formula v = √(GM/r). Substituting the values, we get t = 2.4 hours.
testme

Homework Statement

A bit of asteroid material of mass 2.5 x 10^6 kg orbits around the Earth at an altitude of 2800km. How long does it take the asteroid to complete one orbit.

M asteroid = 2.5 x 10^6 kg
altitude of asteroid = 2800 km
radius Earth = 6.37 x 10^6 m

Fnet = mv^2/r
C = 2∏r
t = d/v

The Attempt at a Solution

r = rEarth + rAsteroid + altitude
r = 9.2 x 10^6 m

C = 2∏r
C = 5.78 x 10^7
C = d

Fg = mv^2/r
mg = mv^2/r
9.8 = v^2/r
9.8 (9.2 x 10^6) = v^2
v = 9500 m/s

t = d / v
t = 5.78 x 10^7 / 9500
t = 6084 s
t = 1.7 h

My friend got 0.29 and the teacher had told him he was right. I didn't have a chance to see what I did wrong so I'm wondering if anyone here can tell me where I made my mistake.

Use Newton's law of gravitation to find the gravitational force when the object is not very close the the surface of the Earth; f = m*g just won't cut it here.

That would be

F = G * M * m/r if I'm not mistaken?

Would F be mg still? Also how does that help me find the velocity?

testme said:
That would be

F = G * M * m/r if I'm not mistaken?

Would F be mg still? Also how does that help me find the velocity?

The r should be squared in your formula.

No, F is not mg. You just wrote the equation for the force due to gravity. Use that where you used mg in your work.

I think I know what you mean, would it be like this.

mv^2/r = (G)(M)(m)/r^2

v^2/r = (G)(M)/ r^2

v^2 = (G)(M)/r

Before I substitute values in, how do I know what to put for M and what m would be? I know that the two m would cancel out and that was the mass of the asteroid, so then M should be the mass of the earth. Generally speaking though, how do I know what mass to use?

Edit:

I tried substituting values in and I got 2.4 h which is still wrong apparently.

Work:

v^2 = GM/r
v^2 = (6.67 x 10^-11)(5.96 x 10^24) / 9.2 x 10^6
V = 6573

t = d/v
t = 5.78 x 10^7 / 6573
t = 8794 s
t = 2.4 h

Last edited:
testme said:
I think I know what you mean, would it be like this.

mv^2/r = (G)(M)(m)/r^2

v^2/r = (G)(M)/ r^2

v^2 = (G)(M)/r

Before I substitute values in, how do I know what to put for M and what m would be? I know that the two m would cancel out and that was the mass of the asteroid, so then M should be the mass of the earth. Generally speaking though, how do I know what mass to use?

That looks better. The small m, which was also used in the expression for the centripetal force, represents the mass of the orbiting body. The large M is the mass of the primary, in this case the mass of the Earth.

Ah, I editted the other post not thinking you had seen my previous post. Anyways..

I tried substituting values in and I got 2.4 h which is still wrong according to what I was told.

Work:

v^2 = GM/r
v^2 = (6.67 x 10^-11)(5.96 x 10^24) / 9.2 x 10^6
V = 6573

t = d/v
t = 5.78 x 10^7 / 6573
t = 8794 s
t = 2.4 h

testme said:
Ah, I editted the other post not thinking you had seen my previous post. Anyways..

I tried substituting values in and I got 2.4 h which is still wrong according to what I was told.

Work:

v^2 = GM/r
v^2 = (6.67 x 10^-11)(5.96 x 10^24) / 9.2 x 10^6
V = 6573

t = d/v
t = 5.78 x 10^7 / 6573
t = 8794 s
t = 2.4 h

1. What is gravitational pull on an asteroid?

Gravitational pull on an asteroid refers to the force of attraction between the asteroid and any other object with mass, such as a planet or a larger asteroid. This force is what keeps the asteroid in orbit around the larger object.

2. How is gravitational pull calculated on an asteroid?

The gravitational pull on an asteroid is calculated using the formula F = G * (m1 * m2)/r^2, where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them. This formula was developed by Sir Isaac Newton.

3. Does the size of an asteroid affect its gravitational pull?

Yes, the size of an asteroid can affect its gravitational pull. Generally, the larger the asteroid, the stronger its gravitational pull will be. However, the distance between the asteroid and other objects also plays a role in the strength of the gravitational pull.

4. How long does it take for an asteroid to complete one orbit?

The time it takes for an asteroid to complete one orbit around another object, such as a planet or the sun, is known as its orbital period. The orbital period of an asteroid can vary greatly depending on its distance from the object it is orbiting and its speed. Some asteroids may take only a few hours to complete an orbit, while others may take several years.

5. Can the gravitational pull on an asteroid change over time?

Yes, the gravitational pull on an asteroid can change over time. This can occur if the asteroid's distance from other objects changes, or if its mass changes due to collisions with other objects or outgassing. Changes in the gravitational pull can also affect the asteroid's orbit and potentially cause it to collide with another object or be ejected from the solar system.

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